I need help with the following problem, thank you!
A random sample of 100 credit card users is to be questioned regarding their satisfaction with their credit card company. For simplification, assume each CC user carries just one card and that the market share percentages are of all CC customers of each respective brand.
Credit Card Market Share %
Visa 53.9
MasterCard 28.9
American Express 13.2
Discover 4.0
a) Propose a procedure for randomly selecting the 100 CC users.
b) For random samples of 100 CC users, what is the expected number of customers who carry Visa? Discover?
c) What is the probability that half or more of the sample carry Visa? American Express?
d) Justify the user of the normal approximation to the binomial of part c.
Tough statistics question (binomial/normal)?
a) I would think a cold calling people would be the best way to get this data. Asking people in a store is subject to the store's CC policy on which cards they accepts. A discover card is not accepted you'd likely not get any discover card carriers in that store.
b) answer is the same as the proportions
c and d)
Let Xb be the number of people who carry Visa. Xb has the binomial distribution with n = 100 trials and success probability p = 0.539
In general, if X has the binomial distribution with n trials and a success probability of p then
P[Xb = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)
for values of x = 0, 1, 2, ..., n
P[Xb = x] = 0 for any other value of x.
To use the normal approximation to the binomial you must first validate that you have more than 10 expected successes and 10 expected failures. In other words, you need to have n * p %26gt; 10 and n * (1-p) %26gt; 10.
Some authors will say you only need 5 expected successes and 5 expected failures to use this approximation. If you are working towards the center of the distribution then this condition should be sufficient. However, the approximations in the tails of the distribution will be weaker especially if the success probability is low or high. Using 10 expected successes and 10 expected failures is a more conservative approach but will allow for better approximations especially when p is small or p is large.
In this case you have:
n * p = 100 * 0.539 = 53.9 expected success
n * (1 - p) = 100 * 0.461 = 46.1 expected failures
We have checked and confirmed that there are enough expected successes and expected failures. Now we can move on to the rest of the work.
If Xb ~ Binomial(n, p) then we can approximate probabilities using the normal distribution where Xn is normal with mean μ = n * p, variance σ² = n * p * (1-p), and standard deviation σ
Xb ~ Binomial(n = 100 , p = 0.539 )
Xn ~ Normal( μ = 53.9 , σ² = 24.8479 )
Xn ~ Normal( μ = 53.9 , σ = 4.984767 )
I have noted two different notations for the Normal distribution, one using the variance and one using the standard deviation. In most textbooks and in most of the literature, the parameters used to denote the Normal distribution are the mean and the variance. In most software programs, the standard notation is to use the mean and the standard deviation.
The probabilities are approximated using a continuity correction. We need to use a continuity correction because we are estimating discrete probabilities with a continuous distribution. The best way to make sure you use the correct continuity correction is to draw out a small histogram of the binomial distribution and shade in the values you need. The continuity correction accounts for the area of the boxes that would be missing or would be extra under the normal curve.
P( Xb %26lt; x) ≈ P( Xn %26lt; (x - 0.5) )
P( Xb %26gt; x) ≈ P( Xn %26gt; (x + 0.5) )
P( Xb ≤ x) ≈ P( Xn ≤ (x + 0.5) )
P( Xb ≥ x) ≈ P( Xn ≥ (x - 0.5) )
P( Xb = x) ≈ P( (x - 0.5) %26lt; Xn %26lt; (x + 0.5) )
P( a ≤ Xb ≤ b ) ≈ P( (a - 0.5) %26lt; Xn %26lt; (b + 0.5) )
P( a ≤ Xb %26lt; b ) ≈ P( (a - 0.5) %26lt; Xn %26lt; (b - 0.5) )
P( a %26lt; Xb ≤ b ) ≈ P( (a + 0.5) %26lt; Xn %26lt; (b + 0.5) )
P( a %26lt; Xb %26lt; b ) ≈ P( (a + 0.5) %26lt; Xn %26lt; (b - 0.5) )
In the work that follows Xb has the binomial distribution, Xn has the normal distribution and Z has the standard normal distribution.
Remember that for any normal random variable Xn, you can transform it into standard units via: Z = (Xn - μ ) / σ
P( Xb ≥ 50 ) =
100
∑ P(Xb = x) = 0.8114108
x = 50
≈ P( Xn ≥ 49.5 )
= P( Z ≥ ( 49.5 - 53.9 ) / 4.984767 )
= P( Z ≥ -0.8826892 )
= 0.8112979
For the AMEX
Xb ~ Binomial(n = 100 , p = 0.132 )
Xn ~ Normal( μ = 13.2 , σ² = 11.4576 )
Xn ~ Normal( μ = 13.2 , σ = 3.384908 )
P( Xb ≥ 50 ) =
100
∑ P(Xb = x) = 1.066959e-18
x = 50
≈ P( Xn ≥ 49.5 )
= P( Z ≥ ( 49.5 - 13.2 ) / 3.384908 )
= P( Z ≥ 10.72407 )
= 0
apricot
Friday, July 31, 2009
Tough statistics question (binomial/normal)?
I need help with the following problem, thank you!
A random sample of 100 credit card users is to be questioned regarding their satisfaction with their credit card company. For simplification, assume each CC user carries just one card and that the market share percentages are of all CC customers of each respective brand.
Credit Card Market Share %
Visa 53.9
MasterCard 28.9
American Express 13.2
Discover 4.0
a) Propose a procedure for randomly selecting the 100 CC users.
b) For random samples of 100 CC users, what is the expected number of customers who carry Visa? Discover?
c) What is the probability that half or more of the sample carry Visa? American Express?
d) Justify the user of the normal approximation to the binomial of part c.
Tough statistics question (binomial/normal)?
a) I would think a cold calling people would be the best way to get this data. Asking people in a store is subject to the store's CC policy on which cards they accepts. A discover card is not accepted you'd likely not get any discover card carriers in that store.
b) answer is the same as the proportions
c and d)
Let Xb be the number of people who carry Visa. Xb has the binomial distribution with n = 100 trials and success probability p = 0.539
In general, if X has the binomial distribution with n trials and a success probability of p then
P[Xb = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)
for values of x = 0, 1, 2, ..., n
P[Xb = x] = 0 for any other value of x.
To use the normal approximation to the binomial you must first validate that you have more than 10 expected successes and 10 expected failures. In other words, you need to have n * p %26gt; 10 and n * (1-p) %26gt; 10.
Some authors will say you only need 5 expected successes and 5 expected failures to use this approximation. If you are working towards the center of the distribution then this condition should be sufficient. However, the approximations in the tails of the distribution will be weaker especially if the success probability is low or high. Using 10 expected successes and 10 expected failures is a more conservative approach but will allow for better approximations especially when p is small or p is large.
In this case you have:
n * p = 100 * 0.539 = 53.9 expected success
n * (1 - p) = 100 * 0.461 = 46.1 expected failures
We have checked and confirmed that there are enough expected successes and expected failures. Now we can move on to the rest of the work.
If Xb ~ Binomial(n, p) then we can approximate probabilities using the normal distribution where Xn is normal with mean μ = n * p, variance σ² = n * p * (1-p), and standard deviation σ
Xb ~ Binomial(n = 100 , p = 0.539 )
Xn ~ Normal( μ = 53.9 , σ² = 24.8479 )
Xn ~ Normal( μ = 53.9 , σ = 4.984767 )
I have noted two different notations for the Normal distribution, one using the variance and one using the standard deviation. In most textbooks and in most of the literature, the parameters used to denote the Normal distribution are the mean and the variance. In most software programs, the standard notation is to use the mean and the standard deviation.
The probabilities are approximated using a continuity correction. We need to use a continuity correction because we are estimating discrete probabilities with a continuous distribution. The best way to make sure you use the correct continuity correction is to draw out a small histogram of the binomial distribution and shade in the values you need. The continuity correction accounts for the area of the boxes that would be missing or would be extra under the normal curve.
P( Xb %26lt; x) ≈ P( Xn %26lt; (x - 0.5) )
P( Xb %26gt; x) ≈ P( Xn %26gt; (x + 0.5) )
P( Xb ≤ x) ≈ P( Xn ≤ (x + 0.5) )
P( Xb ≥ x) ≈ P( Xn ≥ (x - 0.5) )
P( Xb = x) ≈ P( (x - 0.5) %26lt; Xn %26lt; (x + 0.5) )
P( a ≤ Xb ≤ b ) ≈ P( (a - 0.5) %26lt; Xn %26lt; (b + 0.5) )
P( a ≤ Xb %26lt; b ) ≈ P( (a - 0.5) %26lt; Xn %26lt; (b - 0.5) )
P( a %26lt; Xb ≤ b ) ≈ P( (a + 0.5) %26lt; Xn %26lt; (b + 0.5) )
P( a %26lt; Xb %26lt; b ) ≈ P( (a + 0.5) %26lt; Xn %26lt; (b - 0.5) )
In the work that follows Xb has the binomial distribution, Xn has the normal distribution and Z has the standard normal distribution.
Remember that for any normal random variable Xn, you can transform it into standard units via: Z = (Xn - μ ) / σ
P( Xb ≥ 50 ) =
100
∑ P(Xb = x) = 0.8114108
x = 50
≈ P( Xn ≥ 49.5 )
= P( Z ≥ ( 49.5 - 53.9 ) / 4.984767 )
= P( Z ≥ -0.8826892 )
= 0.8112979
For the AMEX
Xb ~ Binomial(n = 100 , p = 0.132 )
Xn ~ Normal( μ = 13.2 , σ² = 11.4576 )
Xn ~ Normal( μ = 13.2 , σ = 3.384908 )
P( Xb ≥ 50 ) =
100
∑ P(Xb = x) = 1.066959e-18
x = 50
≈ P( Xn ≥ 49.5 )
= P( Z ≥ ( 49.5 - 13.2 ) / 3.384908 )
= P( Z ≥ 10.72407 )
= 0
A random sample of 100 credit card users is to be questioned regarding their satisfaction with their credit card company. For simplification, assume each CC user carries just one card and that the market share percentages are of all CC customers of each respective brand.
Credit Card Market Share %
Visa 53.9
MasterCard 28.9
American Express 13.2
Discover 4.0
a) Propose a procedure for randomly selecting the 100 CC users.
b) For random samples of 100 CC users, what is the expected number of customers who carry Visa? Discover?
c) What is the probability that half or more of the sample carry Visa? American Express?
d) Justify the user of the normal approximation to the binomial of part c.
Tough statistics question (binomial/normal)?
a) I would think a cold calling people would be the best way to get this data. Asking people in a store is subject to the store's CC policy on which cards they accepts. A discover card is not accepted you'd likely not get any discover card carriers in that store.
b) answer is the same as the proportions
c and d)
Let Xb be the number of people who carry Visa. Xb has the binomial distribution with n = 100 trials and success probability p = 0.539
In general, if X has the binomial distribution with n trials and a success probability of p then
P[Xb = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)
for values of x = 0, 1, 2, ..., n
P[Xb = x] = 0 for any other value of x.
To use the normal approximation to the binomial you must first validate that you have more than 10 expected successes and 10 expected failures. In other words, you need to have n * p %26gt; 10 and n * (1-p) %26gt; 10.
Some authors will say you only need 5 expected successes and 5 expected failures to use this approximation. If you are working towards the center of the distribution then this condition should be sufficient. However, the approximations in the tails of the distribution will be weaker especially if the success probability is low or high. Using 10 expected successes and 10 expected failures is a more conservative approach but will allow for better approximations especially when p is small or p is large.
In this case you have:
n * p = 100 * 0.539 = 53.9 expected success
n * (1 - p) = 100 * 0.461 = 46.1 expected failures
We have checked and confirmed that there are enough expected successes and expected failures. Now we can move on to the rest of the work.
If Xb ~ Binomial(n, p) then we can approximate probabilities using the normal distribution where Xn is normal with mean μ = n * p, variance σ² = n * p * (1-p), and standard deviation σ
Xb ~ Binomial(n = 100 , p = 0.539 )
Xn ~ Normal( μ = 53.9 , σ² = 24.8479 )
Xn ~ Normal( μ = 53.9 , σ = 4.984767 )
I have noted two different notations for the Normal distribution, one using the variance and one using the standard deviation. In most textbooks and in most of the literature, the parameters used to denote the Normal distribution are the mean and the variance. In most software programs, the standard notation is to use the mean and the standard deviation.
The probabilities are approximated using a continuity correction. We need to use a continuity correction because we are estimating discrete probabilities with a continuous distribution. The best way to make sure you use the correct continuity correction is to draw out a small histogram of the binomial distribution and shade in the values you need. The continuity correction accounts for the area of the boxes that would be missing or would be extra under the normal curve.
P( Xb %26lt; x) ≈ P( Xn %26lt; (x - 0.5) )
P( Xb %26gt; x) ≈ P( Xn %26gt; (x + 0.5) )
P( Xb ≤ x) ≈ P( Xn ≤ (x + 0.5) )
P( Xb ≥ x) ≈ P( Xn ≥ (x - 0.5) )
P( Xb = x) ≈ P( (x - 0.5) %26lt; Xn %26lt; (x + 0.5) )
P( a ≤ Xb ≤ b ) ≈ P( (a - 0.5) %26lt; Xn %26lt; (b + 0.5) )
P( a ≤ Xb %26lt; b ) ≈ P( (a - 0.5) %26lt; Xn %26lt; (b - 0.5) )
P( a %26lt; Xb ≤ b ) ≈ P( (a + 0.5) %26lt; Xn %26lt; (b + 0.5) )
P( a %26lt; Xb %26lt; b ) ≈ P( (a + 0.5) %26lt; Xn %26lt; (b - 0.5) )
In the work that follows Xb has the binomial distribution, Xn has the normal distribution and Z has the standard normal distribution.
Remember that for any normal random variable Xn, you can transform it into standard units via: Z = (Xn - μ ) / σ
P( Xb ≥ 50 ) =
100
∑ P(Xb = x) = 0.8114108
x = 50
≈ P( Xn ≥ 49.5 )
= P( Z ≥ ( 49.5 - 53.9 ) / 4.984767 )
= P( Z ≥ -0.8826892 )
= 0.8112979
For the AMEX
Xb ~ Binomial(n = 100 , p = 0.132 )
Xn ~ Normal( μ = 13.2 , σ² = 11.4576 )
Xn ~ Normal( μ = 13.2 , σ = 3.384908 )
P( Xb ≥ 50 ) =
100
∑ P(Xb = x) = 1.066959e-18
x = 50
≈ P( Xn ≥ 49.5 )
= P( Z ≥ ( 49.5 - 13.2 ) / 3.384908 )
= P( Z ≥ 10.72407 )
= 0
Probabilty help needed?
i have solutions for parts A and C, im hoping you can help me with part B
the question is:
mints have a label weight of 20.4 grams. assume the distribution is N(21.37,0.16)
A) let X denote the weight of a single mint seleted at random, find P(X%26lt;20.857)
i got P(20.857-21.37/0.4) = 0.1003 from normal table
C) let Xbar equal the sample mean of 100 mints selected at random and weighed. find P(21.31%26lt;=Xbar%26lt;=21.39
my answer:
with n=100 the distribution becomes (21.37, 0.16/100)
P(21.31-31.37/ sqrt(.0016)) %26lt;= 21.39-21.37/sqrt(.0016)
= 0.6247 from normal tables. does that seem right?
this is the one i dont understand, part B
B)
100 mints are selceted at random and weighed. let Y equal the number of these mints that weigh less that 20.857 grams. approximate (Y%26lt;=5)
thank you!
Probabilty help needed?
Without doing all of the calculations...
Calculate P(X %26lt; 20.857), which you already did in (A).
Then, use that probability in a binomial distribution modeling.
Reply:I'm not exactly sure myself. but i believe it is saying that out of y/100 of the mints weighs less then 20.857 grams.
Really hoped this helped!
(Sorry if it didn't)
the question is:
mints have a label weight of 20.4 grams. assume the distribution is N(21.37,0.16)
A) let X denote the weight of a single mint seleted at random, find P(X%26lt;20.857)
i got P(20.857-21.37/0.4) = 0.1003 from normal table
C) let Xbar equal the sample mean of 100 mints selected at random and weighed. find P(21.31%26lt;=Xbar%26lt;=21.39
my answer:
with n=100 the distribution becomes (21.37, 0.16/100)
P(21.31-31.37/ sqrt(.0016)) %26lt;= 21.39-21.37/sqrt(.0016)
= 0.6247 from normal tables. does that seem right?
this is the one i dont understand, part B
B)
100 mints are selceted at random and weighed. let Y equal the number of these mints that weigh less that 20.857 grams. approximate (Y%26lt;=5)
thank you!
Probabilty help needed?
Without doing all of the calculations...
Calculate P(X %26lt; 20.857), which you already did in (A).
Then, use that probability in a binomial distribution modeling.
Reply:I'm not exactly sure myself. but i believe it is saying that out of y/100 of the mints weighs less then 20.857 grams.
Really hoped this helped!
(Sorry if it didn't)
Probability answer verification and help?
i have solutions for parts A and C, im hoping you can help me with part B
the question is:
mints have a label weight of 20.4 grams. assume the distribution is N(21.37,0.16)
A) let X denote the weight of a single mint seleted at random, find P(X%26lt;20.857)
i got P(20.857-21.37/0.4) = 0.1003 from normal table
C) let Xbar equal the sample mean of 100 mints selected at random and weighed. find P(21.31%26lt;=Xbar%26lt;=21.39
my answer:
with n=100 the distribution becomes (21.37, 0.16/100)
P(21.31-31.37/ sqrt(.0016)) %26lt;= 21.39-21.37/sqrt(.0016)
= 0.6247 from normal tables. does that seem right?
this is the one i dont understand, part B
B)
100 mints are selceted at random and weighed. let Y equal the number of these mints that weigh less that 20.857 grams. approximate (Y%26lt;=5)
thank you!
Probability answer verification and help?
A is correct, good job.
C you get the proper value, good job. Your writing is a little obscure though. P(21.31%26lt;=Xbar%26lt;=21.39) = Φ((21.39-21.37)/√(0.16/100)) - Φ((21.31-21.37)/√(0.16/100)).
62.47% is correct.
B. You know the probability a mint weighs under 20.857 g is 10% (cf. A).
The probability that 5 or fewer mints in a sample of size 100 weigh under 20.857g is then:
P(Y%26lt;=5) = P(Y=0)+ P(Y=1)+ P(Y=2)+ P(Y=3) +P(Y=4) +P(Y=5)
This is a binomial distribution. P(Y=n) = p^n*(1-p)^(100-n)*100Cn
where p=0.1 (the probability a mint weighs less than 20.857g) and 100Cn = 100!/(n!(100-n)!
The correct result (computed using a VBscript program) yields P(Y%26lt;=5)= 5.7%
You can approximate that result using the normal approximation for the binomial distribution.
The equivalent normal distribution is N(np, (√(np(1-p))²) where n is the size of the sample and p your probability (10%).
It is considered valid when np%26gt;=10. here, np = 10, so we should be ok.
The apporximation is done using a normal distribution of N(10, 3²)
Also, when you consider your upper bound (here you have to find P(Y%26lt;=5)), you do a continuity correction because you are switching from a discrete distribution (binomial) to a continuous distribution (normal). You add 0.5 to your limit.
The approximation becomes: P(y%26lt;=5) = Φ((5.5-10)/3)) = Φ(-1.5) = 6.7% Not an ideal approximation, but not bad either.
the question is:
mints have a label weight of 20.4 grams. assume the distribution is N(21.37,0.16)
A) let X denote the weight of a single mint seleted at random, find P(X%26lt;20.857)
i got P(20.857-21.37/0.4) = 0.1003 from normal table
C) let Xbar equal the sample mean of 100 mints selected at random and weighed. find P(21.31%26lt;=Xbar%26lt;=21.39
my answer:
with n=100 the distribution becomes (21.37, 0.16/100)
P(21.31-31.37/ sqrt(.0016)) %26lt;= 21.39-21.37/sqrt(.0016)
= 0.6247 from normal tables. does that seem right?
this is the one i dont understand, part B
B)
100 mints are selceted at random and weighed. let Y equal the number of these mints that weigh less that 20.857 grams. approximate (Y%26lt;=5)
thank you!
Probability answer verification and help?
A is correct, good job.
C you get the proper value, good job. Your writing is a little obscure though. P(21.31%26lt;=Xbar%26lt;=21.39) = Φ((21.39-21.37)/√(0.16/100)) - Φ((21.31-21.37)/√(0.16/100)).
62.47% is correct.
B. You know the probability a mint weighs under 20.857 g is 10% (cf. A).
The probability that 5 or fewer mints in a sample of size 100 weigh under 20.857g is then:
P(Y%26lt;=5) = P(Y=0)+ P(Y=1)+ P(Y=2)+ P(Y=3) +P(Y=4) +P(Y=5)
This is a binomial distribution. P(Y=n) = p^n*(1-p)^(100-n)*100Cn
where p=0.1 (the probability a mint weighs less than 20.857g) and 100Cn = 100!/(n!(100-n)!
The correct result (computed using a VBscript program) yields P(Y%26lt;=5)= 5.7%
You can approximate that result using the normal approximation for the binomial distribution.
The equivalent normal distribution is N(np, (√(np(1-p))²) where n is the size of the sample and p your probability (10%).
It is considered valid when np%26gt;=10. here, np = 10, so we should be ok.
The apporximation is done using a normal distribution of N(10, 3²)
Also, when you consider your upper bound (here you have to find P(Y%26lt;=5)), you do a continuity correction because you are switching from a discrete distribution (binomial) to a continuous distribution (normal). You add 0.5 to your limit.
The approximation becomes: P(y%26lt;=5) = Φ((5.5-10)/3)) = Φ(-1.5) = 6.7% Not an ideal approximation, but not bad either.
Urgent STAT probability?
On the average, 6.7 cars arrive at the drive-up window of a bank every hour. Define the random
variable X to be the number of cars arriving in any hour.
a. What is the appropriate probability distribution for X? Explain how X satisfies the properties of
the distribution.
b. Compute the probability that exactly 5 cars will arrive in the next hour.
c. Compute the probability that no more than 5 cars will arrive in the next hour.
Urgent STAT probability?
a. This a Poisson distribution. I'll let you explain why. I have to assume you know what a Poisson distribution is and how to apply the formulas.
b. Pr(X=5) = [e^(-6.7)*6.7^5]/5! = 0.13849...
c. Pr(X%26lt;=5) = Pr(X=5) + Pr(X=4) + Pr(X=3) + Pr(X=2) + Pr(X=1) + Pr(X=0)
= e^(-6.7)*[6.7^5/5! + 6.7^4/4! + 6.7^3/3! + 6.7^2/2! + 6.7 + 1]
= 0.340649...
song words
variable X to be the number of cars arriving in any hour.
a. What is the appropriate probability distribution for X? Explain how X satisfies the properties of
the distribution.
b. Compute the probability that exactly 5 cars will arrive in the next hour.
c. Compute the probability that no more than 5 cars will arrive in the next hour.
Urgent STAT probability?
a. This a Poisson distribution. I'll let you explain why. I have to assume you know what a Poisson distribution is and how to apply the formulas.
b. Pr(X=5) = [e^(-6.7)*6.7^5]/5! = 0.13849...
c. Pr(X%26lt;=5) = Pr(X=5) + Pr(X=4) + Pr(X=3) + Pr(X=2) + Pr(X=1) + Pr(X=0)
= e^(-6.7)*[6.7^5/5! + 6.7^4/4! + 6.7^3/3! + 6.7^2/2! + 6.7 + 1]
= 0.340649...
song words
A manufacturing company has 5 identical machines which procduce nails.?
The probability that any machine will break down on any given day is 0.10. Define a random variable x to be the number to machines which will break down in a day. Compute the probability that exact 4 machines will break down.
a. .00032
b. .00038
c. .00040
d. .00049
e. none of the above answers is correct
A manufacturing company has 5 identical machines which procduce nails.?
e. 0.00045 (none of the above)
X follows a binomial distribution with parameters p=0.10 and n=5.
P(X=4) = 5!/(4!1!) 0.10^4 0.90^1 = 0.00045
a. .00032
b. .00038
c. .00040
d. .00049
e. none of the above answers is correct
A manufacturing company has 5 identical machines which procduce nails.?
e. 0.00045 (none of the above)
X follows a binomial distribution with parameters p=0.10 and n=5.
P(X=4) = 5!/(4!1!) 0.10^4 0.90^1 = 0.00045
My son's math - help?
Neither of us could figure these out. I'd like to be able to help him with them and check the answers - could you help with that? It would be very much appreciated.
For 1 and 2, there's a set of cards with the letters G E O M E T R Y on them.
1. You randomly select two cards from those at the right, replacing the first before you select the second. Find the probability you select two E’s.
A. 1/64
B. 1/28
C. 1/16
D. 1/56
2. You randomly select two cards from those at the right, not replacing the first before you select the second. Find the probability you select two E’s
A. 1/64
B. 1/28
C. 1/16
D. 1/56
3. How many ways can Leonardo select 3 movies to rent from the top 10 rentals?
A. 720 ways
B. 240 ways
C. 90 ways
D. 120 ways
4. How many ways can Leonardo select 3 movies to rent from the top 10 rentals and arrange them into a triple feature?
A. 720 ways
B. 240 ways
C. 90 ways
D. 120 ways
COLOR ---- NUMBER OF SHIRTS
Blue _________ 8
Red _________ 7
White _________ 12
Other _________ 3
5. The table shows the shirt colors of a random sample of
students in your school. Find the experimental probability
a student chosen at random is wearing a white shirt.
A. 4/14
B. 2/5
C. 1/10
D. 7/30
6. You want to find out who is likely to win the election for
student council president in your school.Which survey plan
describes the best sample?
F. You survey students in your social studies class.
G. You survey students in Drama Club.
H. You survey students outside the cafeteria during lunch hour.
J. You survey basketball players.
7. From 80,000 jars produced, a manufacturer samples 1,200 and finds 7 of
those defective. Estimate the total number of defective jars.
My son's math - help?
I'm gonna say 1 is B
5 is B as well
6 is H the "lunch hour" one
and the others i've got no idea. i'd have to see the sheet first. some of the info is missing. if that's how they gave it to your on then that school is whack
get him outta there
Reply:1) In "or" statements, add. In "and" statements, multiply.
This problem is an and statement because you want an E and an E. 2/8 times 2/8 = 4/64 = 1/16
2) This problem is different from the first because we don't put the E back into the mix before we choose again. That means that we will now multiply 2/8 times 1/7 (there is only one e and 7 letters left). 2/56 = 1/28
3) We are selecting three movies. Order doesn't matter. We have ten choices for the first pick, nine choices for the second pick, and eight choices for the third pick. 10x9x8 = 720.
4) This time order does matter because it asks how they can be arranged. So we take 720 and divide it by 3x2x1. 720 divided by 6 is 120.
5) 12 out of 30 people are probably wearing a white shirt. This reduces to 2/5
6) F G and J only target a small portion of the population. You are likely to get a better sample outside the cafeteria because, presumably, all the students will be at lunch.
7) if 7 out of 1200 are defective, x out of 80,000 are defective. Write this as two fractions and cross multiply to find out how many of the 80,000 are probably defective.
Reply:1) 1/4 * 1/4 = 1/16
2) 1/4 * 1/7 = 1/28
3) 10 * 9 * 8 = 720
4) 720/6 = 120
5) 12/30 = 2/5
6) Sample outside the cafeteria. It's the most random.
7) 7/1200 * 80000 = 467
Reply:1 = C ( 2/8 * 2/8)
2 = B (2/8 * 1/7)
3 = A
4 = D (3 %26amp; 4 ARE HARD TO EXPLAIN.....)
5 = B (12/30 [ 30 comes from adding the four colors together])
6 = H (most random people -- not a clique)
7 = 467 jars (set up a porportion of 7 over 1200 = x over 80000, cross multiply and get 7 * 80000 = 1200x, simplify to 560000 = 1200x, divide both sides of the equasion by 1200 and get 466.66666 and round up to 467 [because you can't have 2/3 of a jar)
i really hope this helps
Reply:first see how many Es there are for #1. there are 2. and there are 8 letters so for the first card that is 1 out of 8 then you selent the second card and that is 1 out of 8 because you replace the card u took so that is 1 out of 8 and then u multiply them them and that is 1 out of 64. and i think that is how it works but i am not 100% sure.
for #2 u do the same except u do not replace the second card so that would be 1 out of 8 multiplied by 1 out of 7 and that is 1 out of 56.
for #3 there are 10 movies and u need to select 3 so for the first movie there are 10 possible choices and for the second there are 9 possible choices because u already chose 1 and for the third there are eight possible choices because u took 2 out and so basically that is 10x9x8=720
i dont know #4
#5 u just add them all up and then make it into a fraction and then simplify.
#6 is H because it offers a wider variety.
#7 it is just 7 out of 1200 = X out of 80,000.
For 1 and 2, there's a set of cards with the letters G E O M E T R Y on them.
1. You randomly select two cards from those at the right, replacing the first before you select the second. Find the probability you select two E’s.
A. 1/64
B. 1/28
C. 1/16
D. 1/56
2. You randomly select two cards from those at the right, not replacing the first before you select the second. Find the probability you select two E’s
A. 1/64
B. 1/28
C. 1/16
D. 1/56
3. How many ways can Leonardo select 3 movies to rent from the top 10 rentals?
A. 720 ways
B. 240 ways
C. 90 ways
D. 120 ways
4. How many ways can Leonardo select 3 movies to rent from the top 10 rentals and arrange them into a triple feature?
A. 720 ways
B. 240 ways
C. 90 ways
D. 120 ways
COLOR ---- NUMBER OF SHIRTS
Blue _________ 8
Red _________ 7
White _________ 12
Other _________ 3
5. The table shows the shirt colors of a random sample of
students in your school. Find the experimental probability
a student chosen at random is wearing a white shirt.
A. 4/14
B. 2/5
C. 1/10
D. 7/30
6. You want to find out who is likely to win the election for
student council president in your school.Which survey plan
describes the best sample?
F. You survey students in your social studies class.
G. You survey students in Drama Club.
H. You survey students outside the cafeteria during lunch hour.
J. You survey basketball players.
7. From 80,000 jars produced, a manufacturer samples 1,200 and finds 7 of
those defective. Estimate the total number of defective jars.
My son's math - help?
I'm gonna say 1 is B
5 is B as well
6 is H the "lunch hour" one
and the others i've got no idea. i'd have to see the sheet first. some of the info is missing. if that's how they gave it to your on then that school is whack
get him outta there
Reply:1) In "or" statements, add. In "and" statements, multiply.
This problem is an and statement because you want an E and an E. 2/8 times 2/8 = 4/64 = 1/16
2) This problem is different from the first because we don't put the E back into the mix before we choose again. That means that we will now multiply 2/8 times 1/7 (there is only one e and 7 letters left). 2/56 = 1/28
3) We are selecting three movies. Order doesn't matter. We have ten choices for the first pick, nine choices for the second pick, and eight choices for the third pick. 10x9x8 = 720.
4) This time order does matter because it asks how they can be arranged. So we take 720 and divide it by 3x2x1. 720 divided by 6 is 120.
5) 12 out of 30 people are probably wearing a white shirt. This reduces to 2/5
6) F G and J only target a small portion of the population. You are likely to get a better sample outside the cafeteria because, presumably, all the students will be at lunch.
7) if 7 out of 1200 are defective, x out of 80,000 are defective. Write this as two fractions and cross multiply to find out how many of the 80,000 are probably defective.
Reply:1) 1/4 * 1/4 = 1/16
2) 1/4 * 1/7 = 1/28
3) 10 * 9 * 8 = 720
4) 720/6 = 120
5) 12/30 = 2/5
6) Sample outside the cafeteria. It's the most random.
7) 7/1200 * 80000 = 467
Reply:1 = C ( 2/8 * 2/8)
2 = B (2/8 * 1/7)
3 = A
4 = D (3 %26amp; 4 ARE HARD TO EXPLAIN.....)
5 = B (12/30 [ 30 comes from adding the four colors together])
6 = H (most random people -- not a clique)
7 = 467 jars (set up a porportion of 7 over 1200 = x over 80000, cross multiply and get 7 * 80000 = 1200x, simplify to 560000 = 1200x, divide both sides of the equasion by 1200 and get 466.66666 and round up to 467 [because you can't have 2/3 of a jar)
i really hope this helps
Reply:first see how many Es there are for #1. there are 2. and there are 8 letters so for the first card that is 1 out of 8 then you selent the second card and that is 1 out of 8 because you replace the card u took so that is 1 out of 8 and then u multiply them them and that is 1 out of 64. and i think that is how it works but i am not 100% sure.
for #2 u do the same except u do not replace the second card so that would be 1 out of 8 multiplied by 1 out of 7 and that is 1 out of 56.
for #3 there are 10 movies and u need to select 3 so for the first movie there are 10 possible choices and for the second there are 9 possible choices because u already chose 1 and for the third there are eight possible choices because u took 2 out and so basically that is 10x9x8=720
i dont know #4
#5 u just add them all up and then make it into a fraction and then simplify.
#6 is H because it offers a wider variety.
#7 it is just 7 out of 1200 = X out of 80,000.
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