Explore BrainMass
Share

This content was STOLEN from BrainMass.com - View the original, and get the already-completed solution here!

See the attachments.

Exercises 2.16,
2.20,
2.24,
2.26 (b, d)

Exercises 4.10 (a,d),
4.14,
4.20,
4.24,
4.32,
4.42,
4.50.

#### Solution Preview

1. The statistics of the grades of a test are: P15 = 40, P55 = 62, P75 = 81. What percentage of the students had a mark that is:

a) More than 40?

Given that P15 = 40.

Therefore, percentage of the students had a mark that is more than 40 = 100 - 15

= 85%

b) Between 40 and 81?

Given that P15 = 40 and P75 = 81

Therefore, percentage of the students had a mark between 40 and 81 = 75 - 40

= 35%

2. Ann's grade for her physics test is 73%; the class results for the physics test have a mean of 60% and a standard deviation of 3. Ann's grade for her biology test is 69%; the class results for the biology test have a mean of 51% and a standard deviation of 2. Which test result has the higher relative position for Ann?

Let X be the grade for Ann's physics test. We can assume that X is normal with mean µ = 60 and standard deviation  = 3.
We need P (X < 73). Standardizing the variable X using and from standard normal tables, we have,
P (X < 73) = = P (Z < 4.3333) = 0.999992657

Hence we can conclude that the grades of almost all students in the class are less than Ann's grade.
Let Y be the grade for Ann's biology test. We can assume that Y is normal with mean µ = 51 and standard deviation  = 2.
We need P (Y < 69). Standardizing the variable X using and from standard normal tables, we have,
P (Y < 69) = = P (Z < 9) = 1

That is, Ann scored higher than all other students in the class.

The performance of Ann in both tests is more or less the same and she might be the topper in both tests. However biology test result has the higher relative position for Ann.

3. The scores for a test have a normal distribution with a mean of 60 and a standard deviation of 5.

Let X be the scores for the test. We can assume that X is normal with mean µ = 60 and standard deviation  = 5.
Standardizing the variable X using and from standard normal tables, we have,
a) Approximately what percentage of the scores is between 50 and 70?

P (50 < X < 70) = = P (-2 < Z < 2) = 0.9545
Details

Normal Probabilities

Common Data
Mean 60
Standard Deviation 5
Probability for a Range
From X Value 50
To X Value 70
Z Value for 50 -2
Z Value for 70 2
P(X<=50) 0.0228
P(X<=70) 0.9772
P(50<=X<=70) 0.9545

b) Approximately what percentage of the scores is less than 55?

P (X < 55) = = P (Z < -1) = 0.1587
Details

Normal Probabilities

Common Data
Mean 60
Standard ...

#### Solution Summary

The solution determines the normally distributed grades.

\$2.19