Thursday, July 30, 2009

Help with Statistics, probability?

ok so i have this problem i need help figuring out how to solve...my teacher is useless :(





1. A few years ago it was reported that 71.7% of adult Americans were satisfied with the job the nation's major airlines were doing. Suppose 78 adult Americans are selected at random and the number who are satisfied is recorded.





a) Find the probability that exactly 61 of them are satisfied.





b) Find the probability that at most 52 of them are satisfied.





c) Find the probability that at least 62 of them are satisfied.





if anyone could help me i'd greatly appreciate it. i'd appreciate a good explanation too. Thanks!





(On a side note...if anyone happens to know how to work this in minitab let me know :) )

Help with Statistics, probability?
Let Xb be the adults satisfied with the airlines. Xb has the binomial distribution with n = 78 trials and success probability p = 0.717





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 = 78 * 0.717 = 55.926 expected success


n * (1 - p) = 78 * 0.283 = 22.074 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 = 78 , p = 0.717 )


Xn ~ Normal( μ = 55.926 , σ² = 15.82706 )


Xn ~ Normal( μ = 55.926 , σ = 3.978324 )





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











a) Because we are looking for a single point we are only going to use the binomial distribution





P( Xb = 61 ) = 0.04618228








b)





P(Xb ≤ 52 ) =





52


∑ P(Xb = x) = 0.1932628


x = 0





≈ P( Xn %26lt; 52.5 )


= P( Z %26lt; ( 52.5 - 55.926 ) / 3.978324 )


= P( Z %26lt; -0.8611668 )


= 0.1945731








c)





P( Xb ≥ 62 ) =





78


∑ P(Xb = x) = 0.07747627


x = 62





≈ P( Xn ≥ 61.5 )


= P( Z ≥ ( 61.5 - 55.926 ) / 3.978324 )


= P( Z ≥ 1.401093 )


= 0.08059318
Reply:a) 61/78 =X*100=answer


b)52/78= X*100=answer (then the answer needs a less than or equal to sign in front of it)


c)62/78=X*100=answer(then the answer needs a greater than or equal to sign in front of it)





all answers should be rounded to the nearest tenth and are percentages (that is the relevance of the opening percentage)


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