Answer:
The probability that at most 10 of the next 150 items produced are unacceptable is 0.8315.
Step-by-step explanation:
Let <em>X</em> = number of items with unacceptable quality.
The probability of an item being unacceptable is, P (X) = <em>p</em> = 0.05.
The sample of items selected is of size, <em>n</em> = 150.
The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 150 and <em>p</em> = 0.05.
According to the Central limit theorem, if a sample of large size (<em>n</em> > 30) is selected from an unknown population then the sampling distribution of sample mean can be approximated by the Normal distribution.
The mean of this sampling distribution is: 
The standard deviation of this sampling distribution is: 
If 10 of the 150 items produced are unacceptable then the probability of this event is:

Compute the value of
as follows:

*Use a <em>z</em>-table for the probability.
Thus, the probability that at most 10 of the next 150 items produced are unacceptable is 0.8315.