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 \(x=\mu + z\sigma\)

Given \(\mu = 1262\) (mean) and \(\sigma = 118\) (standard deviation).
Substitute the values into the formula: \(x=1262+(-0.58)\times118\).
First, calculate \((-0.58)\times118=-68.44\).
Then, \(x = 1262-68.44=1193.56\).

Answer:

\(1194\)