Question

The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with mean 1261 and a standard deviation of 118. ​(a) Determine the 29th 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:-

Given that,

mean = = 1261

standard deviation = = 118

Using standard normal table,

a ) P(-z < Z < z) = 98​%
P(Z < z) - P(Z < z) = 0.98​
2P(Z < z) - 1 = 0.98​
2P(Z < z ) = 1 + 0.98​
2P(Z < z) = 1.98​
P(Z < z) = 1.98​ / 2
P(Z < z) = 0.99
z =- 2.33 and z =2.33

Using z-score formula,

x = z * +

x = -2.33 * 118 +1261

x = 986.06

x = z * +

x = 2.33 * 118 +1261

x =1535.94

b ) P(Z < z) = 29%

= P(Z < z) = 0.29  

= P(Z < -0.6745 ) = 0.29

z = -0.55

Using z-score formula,

x = z * +

x = -0.55 * 118 +1261

x = 1196.1

29^th percentile = 1196.1

c ) P(Z < z) = 25%

= P(Z < z) = 0.25  

= P(Z < -0.6745 ) = 0.25

z = -0.6745

Using z-score formula,

x = z * +

x = -0.67 * 118 +1261

x = 1181.94

First quartile =Q1 = 1181.94

The z dist'n Third quartile is,

P(Z < z) = 75%

= P(Z < z) = 0.75  

= P(Z < 0.6745 ) = 0.75

z = 0.6745

Using z-score formula,

x = z * +

x = 0.67* 118 + 1261

x = 1340.06

Third quartile =Q3 = 1340.06

IQR = Q3 - Q1

= 1340.06 - 1181.94

=158.12

The interquartile range = 158.12