Question

The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with mean

1263

and a standard deviation of

118.

​(a) Determine the

2929th

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

97​%

of bags.

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

Answer 1

Part a)

P ( X < ? ) = 29% = 0.29

Looking for the probability 0.29 in standard normal table to calculate the critical value Z = -0.55

-0.55 = ( X - 1263 ) / 118
X = 1198.1

P ( X < 198.1 ) = 0.29

Part b)

P ( a < X < b) = 0.97

0.97 / 2 = 0.485

Since 0.5 area in standard normal curve is above and below the mean

a = 0.5 - 0.485 = 0.21

b = 0.5 + 0.485 = 0.985

Looking for the probability to find the critical value in standard normal table

Z = -2.17 and Z = 2.17

-2.17 = ( a - 1263 ) / 118

a = 1006.94

2.17 = ( b - 1263 ) / 118

b = 1519.06

P ( 1006.94 < X < 1519.06 ) = 0.97

Part c)

Interquartile range Q3 - Q1

P ( X < ? ) = 0.25

Looking for the probability to find the critical value in standard normal table Z = -0.67

-0.67 = ( a - 1263 ) / 118

a = 1183.94

Q1 = 1183.94

P ( X < ? ) = 0.75

Looking for the probability to find the critical value in standard normal table Z = 0.67

0.67 = ( b - 1263 ) / 118

b = 1342.06

Q3 = 1342.06

Interquartile range = Q3 - Q1 = 1342.06 - 1183.94 = 158.12