Question

The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with a mean of 1261 chips and a standard deviation of 118 chips. ​(a) Determine the 25th 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?

Answer 1

Here Random variable X:Number of chocolate chips in a bag of chocolate chip cookies.

X follows normal distribution with mean = 1261 and standard deviation = 118

a) 25th percentile:

Z for area = 0.25 is -0.67          (Round to 2 decimal)                  (From statistical table of z values)

      = 1261 - 0.67 *118

      = 1261 - 79.06

      = 1181.94

25 th Percentile = 1181.94

b)

Middle area = 0.95

Left area = (1-0.95) /2 = 0.025

Right area = (1-0.95) /2 = 0.025

Z1 for left area = 0.025 is - 1.96                   (Round to 2 decimal)         (From statistical table of z values)

Z2 for area = 0.95 + 0. 025 = 0.975 is 1.96

        = 1261 - 1.96 * 118

        = 1261 - 231.28

        = 1029.72

        ~ 1030

        = 1261 + 1.96 *118

        = 1261 + 231.28

        = 1492.28

         ~ 1493

The number of chocolate chips in a bag that make up the middle 95​% of bags are 1030 and 1493

c) Interquartile range = Third quartile - first quartile

We calculated first quartile (25th percentile) in part a).

First quartile = 1181.94

Third quartile means 75th percentile

Z for area = 0.75 is 0.67                 (Round to 2 decimal)

      = 1261 + 0.67 *118

      = 1261 + 79.06

      = 1340.06

Interquartile range = 1340.06 - 1181.94

                              = 158.12

Interquartile range = 158.12