Question

 

1.) Find the following areas under the standard normal curve.

a. The area to the left of z = -0.76

b. The area to the right of z = -1.36

c. The area between z = -1.22 and z = 1.33

2.) A large group of students took a test in Finite Math where the grades had a mean of 75 and a standard deviation of 4. Assume that the distribution of these grades is approximated by a normal distribution, and that passing the test is a 65.

a. What percent of students scored higher an 80 or higher?

b.What percent of students failed the test?

c.) What happens when you try to find the percent of students that scored less than a 40?

Answer 1

1.

a. Area to the left of z = -0.76

By using z score table, the area to the left of -0.76 is 0.2236

Table always provides the area to the left of z

Therefore, the area to the left of z = -0.76 is 0.2236

b. Area to the right of z = -1.36

The area to the left of -1.36 according the table is, 0.0869

But the area to the right of -1.36 is nothing but 1 - area to the left of -1.36 which is 1 - 0.0869 = 0.9131

Therefore, area to the right of z = -1.36 is 0.9131

c. Area between z = -1.22 and z = 1.33

The area to the left of z = -1.22 using z score table is 0.1112 and the area to the left of z = 1.33 is 0.9082

To find between area just subtract small area from large that is 0.9082 - 0.1112 which is 0.7970

Therefore, the area between z = -1.22 and z = 1.33 is 0.7970

2. Given:

The distribution of grade student get in test have normal distribution with mean 75 and standard deviation is 4

a. Percent of student score 80 or higher, that is first find

First find the z score for 80

The formula of z score is,

The probability for z = 1.25 using z score table is 0.8944, this is the less than probability.

To get more than just subtract it from 1, that is 1 - 0.8944 = 0.1056

0.1056 * 100 = 10.56%

Therefore, 10.56% of students scored 80 or higher.

b. Percent of student failed test

Test requires 65 to pass

If student get less than 65 then he failed the test.

That is P(X < 65), the z score for 65 is

That is P(X < 65) becomes P(Z < -2.5)

The probability for z score -2.5 using table is, 0.0062

0.0062 * 100 = 0.62%

Therefore, 0.62% student will fail the test.

c. Percent of students that scored less than 40

That is P(X < 40)

The z score for x = 40 is,

That is P(X < 40) becomes P(Z < -8.75)

The probability for z score -8.75 is 0 that is negligible

40 is too far from the mean, so the probability for this is approximately 0

That means 0% of the students scored less than a 40.