Standard deviation & probability?

Shoe store records show 30% of customers use credit card to make payment.





Q1.a. Estimate probability that of next 20 purchases, @ least 5 will use credit card.





b. assume 200 people make purchases each day. Estimate mean %26amp; standard deviation of no. customers who use credit card.





c. Find probability that fewer than 50 of 200 purchases on one day are made with credit card.

Standard deviation %26amp; probability?
problem a:





Is a binomial distribution with the p = 0.30, q = 0.70, n = 20





P(X%26gt;=5) = 1 - P(X%26lt;=4) = 1 - 0.2374 = 0.7626


(used a table)





Problem b:





mean is N*P = 200* 0.30 = 60


standard deviation is sqrt(N*P*Q) = 6.48





Problem c:


We should consider using the Normal Approximation to the Binomial Distribution, but first we must check that it applies since 0.30 is fairly far away from the ideal case of p = 0.5.





That rule is that both np and n(1-p) are greater than 5 and is satisfied.





Z = (X - np)/sqrt(npq) is near a standard normal distribution.





We are looking for the Z value, but we use the method with the continuity correction, where the integer 50 is approximated as 50.5 (see your text).





Z = (50.5 - 60)/6.48 = - 1.47





So we look in a standard normal table to find the area under the curve to the left of -1.47, that is 0.0708 round to 0.07





Please check for errors.
Reply:Q1(a)





This is a binomial distribution application for a Bernoulli trial.


let p = probability of a customer using credit card = 0.3


q = 1-p=0.7 (prob. of not using a credit card).


P(x %26gt;=5) = 1 - P(x%26lt;=4) = 1 - 20!/[(20-4)!4!] * (0.3)^4 (0.7)^(20-4)


= 1 - 20.19.18.17/(4.3.2.1)*(0.3^4)*(0.7^16) = 0.87.





Q1(b)


Mean = np = 200*0.3=60.


Variance = npq = 200*0.3*0.7 = 42.


Standard dev = sqrt(variance)=6.48.





(C)


P(X%26lt;=50) = 200!/(150! 50!)*(0.3)^4*(0.7)^150. This is hard to evaluate. Use the asymptotic assumption of a gaussian distribution.





z = (x-mu)/sigma = (50-60)/6.48 = -1.54


P(z %26lt;= -1.54) = 0.06. (from normal tables).





See,


http://mathworld.wolfram.com/BinomialDis...
Reply:For C you use the normal approximation to the binomial distribution. Since you have a large number of trials this distribution will approach normalcy.





Mean=np=60


variance=42





P(n%26lt;50) is equivalent to P(z%26lt;(49.5-60)/6.48))





Edit: You use 49.5 due to the fact that you are approximating a discrete distribution (binomial) with a continuous one (standard normal).