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 27th 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

Solution:-

Given that,

mean = =

standard deviation = = 118

Using standard normal table,

a ) P(Z < z) = 27%

P(Z < z) = 0.27

P(Z < - 0.61) = 0.27

z = - 0.61

Using z-score formula,

x = z * +

x = - 0.61 *118 +1261

x = 1189.02

The 27th percentile = 1189.02

b ) P(-z < Z < z) = 97​%
P(Z < z) - P(Z < z) = 0.97​
2P(Z < z) - 1 = 0.97​
2P(Z < z ) = 1 + 0.97​
2P(Z < z) = 1.97​
P(Z < z) = 1.97​ / 2
P(Z < z) = 0.985
- z = -2.17 and z = 2.17

The z dist'n First quartile is,

Using z-score formula,

x = z * +

x = -2.17 *118 +1261

x = 1004.94

x = z * +

x = 2.17 *118 +1261

x = 1517.06

c ) P(Z < z) = 25%

= P(Z < z) = 0.25  

= P(Z < -0.6745 ) = 0.25

z = -0.67

Using z-score formula,

x = z * +

x = -0.67 *118 +1261

x = 1181.94

First quartile =Q1 = 1181.94

P(Z < z) = 75%

= P(Z < z) = 0.75  

= P(Z < 0.6745 ) = 0.75

z = 0.67

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

IQR = 158.12