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 117 chips. (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

Given that,

mean = $\mu$ = 1261

standard deviation = $\sigma$ = 117

Using standard normal table,

P(Z < z) = 28%

=(Z < z) = 0.28

= P(Z < -0.58 ) = 0.28

z = -0.58

Using z-score formula  

x = z $\sigma$ + $\mu$

x = -0.58*117+1261

x = 1193.14

(B)

middle 98% of score is

P(-z < Z < z) = 0.98

P(Z < z) - P(Z < -z) = 0.98

2 P(Z < z) - 1 = 0.98

2 P(Z < z) = 1 + 0.98= 1.98

P(Z < z) = 1.98 / 2 = 0.99

P(Z < 2.33) = 0.99

z  ±2.33

Using z-score formula  

x= z * $\sigma$ + $\mu$

x =- 2.33*117+1261

x= 988.39

z = 2.33

Using z-score formula  

x= z * $\sigma$ + $\mu$

x= 2.33*117+1261

x= 1533.61

(C)

Using standard normal table,

The z dist'n First quartile is,

P(Z < z) = 25%

= P(Z < z) = 0.25  

= P(Z < -0.6745 ) = 0.25

z = -0.6745

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = -0.6745 * 117+1261

x = 1182.0835

First quartile =Q1 = 1182.0835

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 * $\sigma$ + $\mu$

x = 0.6745 * 117+1261

x = 1339.9165

Third quartile =Q3 = 1339.9165

IQR = Q3 - Q1=1339.9165 -1182.0835 =157.8330