Question

Overview

In this assignment you will apply what you have learned about standard deviation and normal curves to solve a number of problems, including finding the area under selected sections of a normal curve using z-scores, then solve a problem using normal distributions.

Part 1:

The weights of all one hundred (100) 9th graders at a school are measured, and it is found that the mean of all the measurements is 100 lbs., with a standard deviation of 15 lbs.   Explain how you would use this information to determine the percentage of students who weighed between 85 lbs and 115 lbs. Make sure your explanation includes the use of z-scores.

Part 2:

Using the example in part 1 above, answer the following questions in the same document you used to answer the questions from part 1. Make sure you show all the work required by the questions.

• What is the percentage of students who weighed between 85 lbs and 115 lbs? (Show calculations)
• What is the z-score of a student who weighed 105 lbs? (Show calculations)
• If a student weighed only 60 lbs, how many students out of the one hundred 9th graders would weigh more? (Hint: calculate the percentage 1st)

Part 3:

The following year, the 9th grade students are weighed again, and this time it is found that the mean of all weights is 108 lbs, and the standard deviation is 17 lbs. Answer the following questions, using the same document you used in parts 1 and 2 to show how you arrived at your answers.

Compare two students, one from the first class who weighed 98 lbs, and one from the second class who weighed 100 lbs.

• Which of the two students was heavier relative to their class? (Hint:compare z-scores)
• What percentage of students in the first class were heavier than the 98 lb. student?
• What percentage of students in the second class were heavier than the 100 lb. student?

Answer 1

Part 1)

Let x be the weight of the student in the 9th graded.

x follows normal distribution with µ = 100 lbs and σ = 15 lbs

We are asked to find P ( 85 ≤ x ≤ 115 )

So first we need to find z score for x = 115 using z score formula and then area under the corresponding z score using z score table

Then we have to find z score for x = 85 using z score formula and then area under the corresponding z score using z score table.

Then we will subtract the area under both z scores to get the area between 85 and 115 , then we will express it in %

Part 2)  

a) P ( 85 ≤ x ≤ 115 ) = P( x ≤ 115 ) - P( x ≤ 85 )

=     

= P( z < 1 ) - P( z < -1 )

= 0.8413 - 0.1587 ----- ( from z score table )

=0.6827 or 68.27%

So percentage of students who weighed between 85 lbs and 115 lbs is 68.27%

b) z score for x = 105

z =   = 0.33

c) P( x > 60 ) = = P ( z > -2.67 )

= 1- P( z < -2.67 )

= 1 - 0.0038

= 0.9962

So 100*0.9962 = 99.62 ~ 100

Therefore all 100 students out of the one hundred 9th graders would weigh more than 60lbs

Part 3)

a) We are given for second class  µ = 108 lbs and σ = 17 lbs

Student from the first class who weighed 98 lbs has z score

z = = -0.13

Student from the second class who weighed 100 lbs has z score

z = = -0.47

-0.13 is greater than -0.47 , therefore  first class student who weighed 98 lbs was heavier.

b) P( x₁ > 98 ) = = P( z > -0.13 ) = 1 - P( z < -0.13 )

= 1 - 0.4483

= 0.5517

So percentage of students in the first class were heavier than the 98 lb. student is 55.17%

c)

P( x₂ > 100 ) = = P( z > -0.47 ) = 1 - P( z < -0.47 )

= 1 - 0.3192

= 0.6808

So percentage of students in the second class were heavier than the 100 lb. student is 68.08%