Question

A set of final examination grades in a calculus course wasfound to be normally distributed with a mean of 69 and a standarddeviation of 9.

a. what is the probality of getting a grade of 91or less on this exam?

b. What percentage of students scored between 65 and89?

c. What percentage of students scored between 81 and89?

d. Only 5% of the students taking the test scored higherthan what grade?

Answer 1

Concepts and reason

Normal distribution: Normal distribution is a continuous distribution of data that has the bell-shaped curve. The normally distributed random variable x has meanand standard deviation.

Also, the standard normal distribution represents a normal curve with mean 0 and standard deviation 1. Thus, the parameters involved in a normal distribution are mean and standard deviation.

Standardized z-score: The standardized z-score represents the number of standard deviations the data point is away from the mean.

• If the z-score takes positive value when it is above the mean (0).

• If the z-score takes negative value when it is below the mean (0).

Fundamentals

Let , then the standard z-score is found using the formula given below:

Where X denotes the individual raw score, denotes the population mean and denotes the population standard deviation.

(a)

The probability of getting a grade of 91 or less on this exam is obtained below:

Let X denotes the random variable which follows normal distribution with mean 69 and the standard deviation of 9. That is, .

The required probability is obtained below:

From the “standard normal table”, the area to the left of is 0.9930

(b)

The percentage of students scored between 65 and 89 is obtained below:

From the information, it is clear that a final examination grades between 65 and 89.

Let X denotes the final examination grades which follows normal distribution with mean 69 and the standard deviation of 9. That is, .

The required probability is,

From the “standard normal table”, the area to the left of is 0.9868 and the area to the left of is 0.3299.

In percentage of students scored between 65 and 89 is,

(c)

The percentage of student scored between 81 and 89 is obtained below:

Let X denotes the final examination grades which follows normal distribution with mean 69 and the standard deviation of 9. That is, .

The required probability is obtained below:

From the “standard normal table”, the area to the left of is 0.9868 and the area to the left of is 0.9082.

In percentage of students scored between 81 and 89 is,

(d)

The 5% of the students taking the test scored higher than grade is obtained below:

The value of z for the 5% of students taking the test scored higher than grade is obtained as shown below:

Consider,

Procedure for finding the value z is listed below:

1.From the table of standard normal distribution, locate the probability value of 0.95.

2.Move left until the first column is reached. Note the value as 1.6.

3.Move upward until the top row is reached. Note the value as 0.04.

4.The intersection of the row and column values gives the area to the left of$z=1.64 $

The required amount is,

Ans: Part a

The value of is 0.9930.

Part b

The percentage of students scored between 65 and 89 is 65.69%.

Part c

The percentage of student scored between 81 and 89 is 7.86%.

Part d

The 5% of the students taking the test scored higher than grade is 83.76.