Question

The manager of store A is aware that waiting times are much longer if the customer makes an order with a special request. From past experience this occurs 20% of the time. If we monitor the next 10 customers,

(a) What is the probability that half or more of these customers make a special request?

(b) What assumptions do you need to make to find this probability? Out of the next 100 customers,

(c) There is a probability of 90% that the number of customers make a special request equals or exceeds what value?  

Answer 1

(a) Here we have two possibilities that,customer makes an order with a special request or not.

Hence we have two out comes. means binomial process

where probability (p) that customer makes an order with special request= 20%=0.2

we want two check for probability that half or more of these customer makes a special request.

Hence n=10

using pdf od binomial distribution,

p(x)= ; n=0,1,2,3,.......................

= 0 ;Otherwise.

since we want to find probability that, half or more of these customers make a special request.p(x>=5),with p=0.2 and n=10.

required probability=p(x=5)+p(x=6)+p(x=7)+p(x=8)+p(x=9)+p(x=10)

putting values of p, n and x in pdf of binomial distribution,

required probability=

required probability = 0.03279

(b) next we want this to be for 100 customer. means out n=100 which is large number.

we assume that our sample size is large. and for large sample size binomial distribution tends to Normal distribution( Using Central Limit Theorm).

Hence X ~ N(np, np(1-p))

X~ N(20,16)

(c) There is a probability of 90% that the number of customers make a special request equals or exceeds what value?

Which means we want to find probability that,

P(X>=x)=0.9

X~N(Mean=20, variance= 16)

hence, P(X>=x)=0.9

=0.9

Using normal tables search for probability 0.9 in table and that value will give you z value =1.29

solving this you will get, x=20.519 , x 21 (since here we are counting number of customers)