In some bank, the time it takes from the moment a customer arrives until a clerk is available is distributed normally with $\mu=15,\sigma=2$

a. What's the probability for the next client to wait more than 18 minutes?

b. What's the prob. that the average waiting time of the next 100 customers will be greater than 15.1 minutes (assuming that the waiting times are independent)

c. The bank management hired a new manager for customer service and he claims that since he arrived at this job, the time it takes to be treated by a clerk decreased. To check his claim he measured the waiting time of 30 random customers and got an average of 14.3 minutes. Assuming that the s.d. hasn't changed - Check if his claim is true, with confidence level of a=0.1.

My answer

a. P(X>18)=(normalizing) P(Z>1.5)=0.0668

b. $P(\bar X>15.1)=P(Z>0.5)=0.3085$

c. $H_0:\mu=15, H_A:\mu<15$ so $CI=(-\infty,15-Z_0.99 \frac 2{\sqrt{30}}]=(-\infty,14.15]$ but 14.13<14.15 so H_0 is true and not rejected.

Also calculating the p-value: $P_{H_0}(\bar X<14.3)=P(Z<-1.91)=0.0281>0.01 $therefore $H_0$ is not rejected.

Since this is the first time I've been doing it on my own I'd love if someone take a look, and confirm that it's correct (or not?)


1 Answer 1


Since this is self-study I'll just give hints. In b, check your division. in c, check your wording

  • $\begingroup$ the fault in c is that it's not that H0 is true, it's just not false - right? $\endgroup$
    – jreing
    Jan 25, 2014 at 14:40
  • $\begingroup$ Yes, that's right. We "fail to reject" the null. $\endgroup$
    – Peter Flom
    Jan 26, 2014 at 10:12

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