# The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with a mean of 1261chips 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 97​% of bags.​(c) What is the interquartile range of the number of chocolate chips in a bag of chocolate chip​ cookies?

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 97​% of bags.
​(c) What is the interquartile range of the number of chocolate chips in a bag of chocolate chip​ cookies?
check_circle

Step 1

Solution:

a.The 28th percentile for the number of chocolate chips in a bag is obtained below:

From the given information, mean value is 1,261, standard deviation value is 117. Let the random variable X be the number of chocolate chips in a bag of chocolate chip cookie  follows normal distribution with population mean as 1,261 chips and standard deviation score as 117 chips.

From Excel, NORM.INV(0.28,1261,117)=1192.81

Step 2

b.The number of chocolate chips in a bag that make up the middle 97​% of bags is obtained below:

The lower tail value is obtained below:

The lower tail probability value=(1-0.97)/2=0.015

From Excel, NORM.INV(0.015,1261,117)=1007.1

The upper tail value is obtained below:

The upper tail probability value=0.97+0.015=0.985

From Excel, NORM.INV(0.985,1261,117)=1514.9

The number of chocolate chips in a bag that make up the middle 97% of bags is(1007.1,1514.9)

Step 3
1. The interquartile range of the number of chocolate chips in a bag of chocolate chip​ cookies is obtained below:

The third quartile probability value=0.75

The third quartile value is,

From Excel, NORM.INV(0.75,1261,117)=1339.92.

The first quartile probability value=0.25

The first quartile value is,

From Excel, NORM.INV(0.25,1261,117)=1182.08.

The required inter quartile range is,

Q3-Q1=1339.92-1182.08=15...

### Want to see the full answer?

See Solution

#### Want to see this answer and more?

Solutions are written by subject experts who are available 24/7. Questions are typically answered within 1 hour.*

See Solution
*Response times may vary by subject and question.
Tagged in

### Other 