The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with a mean of 1262 chips and a standard deviation of 118 chips. Required:
a. Determine the 26th percentile for the number of chocolate chips in a bag.
b. Determine the number of chocolate chps in a bag that make up the middle 96% of bags.

(a) The 26th percentile for the number of chocolate chips in a bag is 1185

(b) The number of chocolate chips in a bag that makes up the middle 96% of the bags is between 1020 and 1504

Step-by-step explanation:

From the question, we have the following values:

μ =1262 and σ =118

(a) Let the value of x represents the 26th percentile. So the 26th percentile means 26% data is less than x. We can use the standard normal table to get the particular z-value that corresponds to this percentile.

P( Z<-0.65 )=0.2578 which is approximately 0.26

So for 26th percentile z-score will be -0.65.

Mathematically;

z-score = (x-mean)/SD

-0.65 = (x-1262)/118

-76.7 = x -1262

x = 1262-76.7 = 1185.3

This value is approximately 1,185

(b) Using a graph of standard normal distribution curve, if middle is 96% , then at both tails 2% each.

From z-table, we can find the closest probability;

P(-2.05<z<2.05) = 0.96

So we have two x values to get from the individual z-scores

-2.05 = (x-1262)/118

x = 1020(approximately)

For 2.05, we have

2.05 = (x-1262)/118

x = 1262 + 2.05(118) = 1504 (approximately)

b

step-by-step explanation: