Question

Asked Oct 25, 2019

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.

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?

Step 1

**Solution:**

**a.The 28 ^{th} 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

**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...**

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