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)

add all the numbers and check the ones place.

the answer is 486. as the ones place is more than 4, the estimation will be the next 10 ie, 490

c

step-by-step explanation: