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 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 96​% 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 = = 1261

standard deviation = = 118

(a)Using standard normal table,

P(Z < z) = 29%

= P(Z < z) = 0.29  

= P(Z < - 0.55 ) = 0.29

z = - 0.55

Using z-score formula  

x= z * +

x= - 0.55 *118+1261

x= 1196.1

(b)middle 96% of score is

P(-z < Z < z) = 0.96

P(Z < z) - P(Z < -z) = 0.96

2 P(Z < z) - 1 = 0.96

2 P(Z < z) = 1 + 0.96 = 1.96

P(Z < z) = 1.96 / 2 = 0.98

P(Z <2.05 ) = 0.98

z  ±2.05

z= - 2.05

Using z-score formula  

x= z * +

x=2.05 *118+1261

x= 1019.1

z = 2.05

Using z-score formula  

x= z * +

x=2.05 *118+1261

x= 1502.9