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

apricot