Question

Choose the best answer. The distribution of grade point averages for a certain college is approximately Normal with a mean of 2.5 and a standard deviation of 0.6. Within which of the following intervals would we expect to find approximately 81.5% of all GPAs for students at this college?

(a) (0.7, 3.1)
(b) (1.3, 3.7)
(c) (1.9, 3.7)
(d) (1.9, 4.3)
(e) (0.7, 4.3)

Answer 1

Within (1.9 , 3.7) intervals we would expect to find approximately 81.5% of all GPA's for students at this college.

Explanation:

We are given that the distribution of grade point averages for a certain college is approximately Normal with a mean of 2.5 and a standard deviation of 0.6.

Let X = distribution of grade point averages for a certain college

SO, X ~ Normal(
`\mu=2.5,\sigma^(2) =0.6^(2)`)

The z-score probability distribution for normal distribution is given by;

Z =
`(X-\mu)/(\sigma)` ~ N(0,1)

where,
`\mu` = population mean = 2.5

`\sigma` = standard deviation = 0.6

Now, to find that within which interval we expect to find approximately 81.5% of all GPA's for students at this college, we will find respective probabilities for given intervals.

(a) P(0.7 < X < 3.1) = P(X < 3.1) - P(X
`\leq` 0.7)

P(X < 3.1) = P(
`(X-\mu)/(\sigma)` <
`(3.1-2.5)/(0.6)` ) = P(Z < 1) = 0.84134

P(X
`\leq` 0.7) = P(
`(X-\mu)/(\sigma)`
`\leq`
`(0.7-2.5)/(0.6)` ) = P(Z
`\leq` -3) = 1 - P(Z < 3)

= 1 - 0.99865 = 0.00135

Therefore, P(0.7 < X < 3.1) = 0.84134 - 0.00135 = 0.8399

This means that in this interval we are able to find approximately 83.99% of all GPA's for students at this college.

(b) P(1.3 < X < 3.7) = P(X < 3.7) - P(X
`\leq` 1.3)

P(X < 3.7) = P(
`(X-\mu)/(\sigma)` <
`(3.7-2.5)/(0.6)` ) = P(Z < 2) = 0.97725

P(X
`\leq` 1.3) = P(
`(X-\mu)/(\sigma)`
`\leq`
`(1.3-2.5)/(0.6)` ) = P(Z
`\leq` -2) = 1 - P(Z < 2)

= 1 - 0.97725 = 0.02275

Therefore, P(1.3 < X < 3.7) = 0.97725 - 0.02275 = 0.9545

This means that in this interval we are able to find approximately 95.45% of all GPA's for students at this college.

(c) P(1.9 < X < 3.7) = P(X < 3.7) - P(X
`\leq` 1.9)

P(X < 3.7) = P(
`(X-\mu)/(\sigma)` <
`(3.7-2.5)/(0.6)` ) = P(Z < 2) = 0.97725

P(X
`\leq` 1.9) = P(
`(X-\mu)/(\sigma)`
`\leq`
`(1.9-2.5)/(0.6)` ) = P(Z
`\leq` -1) = 1 - P(Z < 1)

= 1 - 0.84134 = 0.15866

Therefore, P(1.9 < X < 3.7) = 0.97725 - 0.15866 = 0.8186

This means that in this interval we are able to find approximately 81.86% of all GPA's for students at this college.

(d) P(1.9 < X < 4.3) = P(X < 4.3) - P(X
`\leq` 1.9)

P(X < 4.3) = P(
`(X-\mu)/(\sigma)` <
`(4.3-2.5)/(0.6)` ) = P(Z < 3) = 0.99865

P(X
`\leq` 1.9) = P(
`(X-\mu)/(\sigma)`
`\leq`
`(1.9-2.5)/(0.6)` ) = P(Z
`\leq` -1) = 1 - P(Z < 1)

= 1 - 0.84134 = 0.15866

Therefore, P(1.9 < X < 4.3) = 0.99865 - 0.15866 = 0.8399

This means that in this interval we are able to find approximately 83.99% of all GPA's for students at this college.

(e) P(0.7 < X < 4.3) = P(X < 4.3) - P(X
`\leq` 0.7)

P(X < 4.3) = P(
`(X-\mu)/(\sigma)` <
`(4.3-2.5)/(0.6)` ) = P(Z < 3) = 0.99865

P(X
`\leq` 0.7) = P(
`(X-\mu)/(\sigma)`
`\leq`
`(0.7-2.5)/(0.6)` ) = P(Z
`\leq` -3) = 1 - P(Z < 3)

= 1 - 0.99865 = 0.00135

Therefore, P(0.7 < X < 4.3) = 0.99865 - 0.00135 = 0.9973

This means that in this interval we are able to find approximately 99.73% of all GPA's for students at this college.

From all these it is clear that the interval (1.9 , 3.7) is expected to find approximately 81.5% of all GPA's for students at this college.