Question

The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with mean 1262 and a standard deviation of 117. ​(a) Determine the 28th 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 98​% of bags. ​(c) What is the interquartile range of the number of chocolate chips in a bag of chocolate chip​ cookies?

Answer 1

Solution :

mean = = 1262

standard deviation = = 117

Using standard normal table,

a ) The 28^th percentile

P(Z < z) = 28%

P(Z < z) = 0.28

P(Z < - 0.5828 ) = 0.28

z = -0.58

Using z-score formula,

x = z * +

x = - 0.58 * 117 + 1262

= 1194.14

The 28^th percentile = 1194.14

b ) The 98^th percentile

P(Z < z) = 98%

P(Z < z) = 0.98

P(Z < 2.054 ) = 0.98

z =2.05

Using z-score formula,

x = z * +

x = 2.05 * 117 + 1262

= 1501.85

The 98^th percentile = 1501.85

​c )  The interquartile range = Q₃ - Q₁

= 1501.85 - 1194.14

= 307.71

The interquartile range = 307.71