Question

the number of chocolate chips in an 18 - ounce bag of chips ahoy! chocolate chip cookies is approximately normally distributed, with a mean of 1262 chips and a standard deviation of 118 chips, according to a study by cadets of the u.s. air force academy.
(a) determine the 28th percentile for the number of chocolate chips in an 18 - ounce bag of chips ahoy! cookies.
(b) determine the number of chocolate chips in a bag of chips ahoy! that make up the middle 98% of bags.
(c) what is the interquartile range of the number of chips in chips ahoy! cookies?
(a) the 28th percentile for the number of chocolate chips in an 18 - ounce bag of chips ahoy! cookies is □ chocolate chips.
(round to the nearest whole number as needed.)

Explanation:

Step1: Find the z - score for the 28th percentile

Using a standard normal distribution table or calculator, the z - score \(z\) corresponding to a cumulative probability of \(0.28\) is approximately \(z=- 0.58\).

Step2: Use the z - score formula to find the value of \(x\)

The z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 1262\) (mean) and \(\sigma=118\) (standard deviation).
We want to solve for \(x\):

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Answer:

\(1194\)