The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with mean of 1262 and a standard deviation of 118. Determine the 26th percentile for the number of chocolate chips in a bag. ​(b) Determine the number of chocolate chips in a bag that make up the middle 95% of bags. ​(c) What is the interquartile range of the number of chocolate chips in a bag of chocolate chip​ cookies?

a) 1186

b) Between 1031 and 1493.

c) 160

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean and standard deviation , the z-score of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Normally distributed with mean of 1262 and a standard deviation of 118.

This means that

a) Determine the 26th percentile for the number of chocolate chips in a bag. ​

This is X when Z has a p-value of 0.26, so X when Z = -0.643.

(b) Determine the number of chocolate chips in a bag that make up the middle 95% of bags.

Between the 50 - (95/2) = 2.5th percentile and the 50 + (95/2) = 97.5th percentile.

2.5th percentile:

X when Z has a p-value of 0.025, so X when Z = -1.96.

97.5th percentile:

X when Z has a p-value of 0.975, so X when Z = 1.96.

Between 1031 and 1493.

​(c) What is the interquartile range of the number of chocolate chips in a bag of chocolate chip​ cookies?

Difference between the 75th percentile and the 25th percentile.

25th percentile:

X when Z has a p-value of 0.25, so X when Z = -0.675.

75th percentile:

X when Z has a p-value of 0.75, so X when Z = 0.675.

IQR:

1342 - 1182 = 160

yes

step-by-step explanation: