Either a delivery is on time (<= 10 minutes), or it isn't, so assume this is a binomial distribution with n=100, p=.9, q=1-p=.1
Then, for a binomial distribution,
mu=np=90, and s=SQRT(npq)=SQRT(100*.9*.1)=3
The sampling distribution of XBAR is a t distribution (because we used a sample standard deviation s), but because n is large we can approximate it with a normal distribution.
We want P[ z = (XBAR-mu)/(s/SQRT(n)) < critical value of z ] = .90
The critical value of z is 1.282 from normal table.
Is (XBAR-mu)/(s/SQRT(n)) < 1.282 ?
XBAR = .82 < 1.282 * (s/SQRT(n)) + mu
.82 < 1.282 * (3/SQRT(100)) + .82 = 1.2046
So since .82 is less than 1.2046, we conclude that less than 90% of orders are delivered in less than 10 minutes.
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