Question

2. The Welcher Adult Intelligence Test Scale is composed of a number of subtests. On one subtest, the raw scores have a mean of 35 and a standard deviation of 6. Assuming these raw scores form a normal distribution: a) What number represents the 65th percentile (what number separates the lower 65% of the distribution)? 37.31 b) What number represents the 90th percentile? 42.71 c) What is the probability of getting a raw score between 28 and 38? 57% d) What is the probability of getting a raw score between 41 and 44? 9

Answer 1

a) 37.31 b) 42.70 c) 0.57 d) 0.09

Step-by-step explaanation:

We are regarding a normal distribution with a mean of 35 and a standard deviation of 6, i.e.,
`\mu` = 35 and
`\sigma` = 6. We know that the probability density function for a normal distribution with a mean of
`\mu` and a standard deviation of
`\sigma` is given by

`f(x) = (1)/(√(2\pi)\sigma)\exp[-((x-\mu)^(2))/(2\sigma^(2))]`

in this case we have

`f(x) = (1)/(√(2\pi)6)\exp[-((x-35)^(2))/(2(6^(2)))]`

Let
`X` be the random variable that represents a row score, we find the values we are seeking in the following way

a) we are looking for a number
`x_(0)` such that

`P(X\leq x_(0))` =
`\int\limits^{x_(0)}_(-\infty) {f(x)} \, dx = 0.65`, this number is
`x_(0)`=37.31

you can find this answer using the R statistical programming languange and the instruction qnorm(0.65, mean = 35, sd = 6)

b) we are looking for a number
`x_(1)` such that

`P(X\leq x_(1))` =
`\int\limits^{x_(1)}_(-\infty) {f(x)} \, dx = 0.9`, this number is
`x_(1)`=42.70

you can find this answer using the R statistical programming languange and the instruction qnorm(0.9, mean = 35, sd = 6)

c) we find this probability as

`P(28\leq X\leq 38)`=
`\int\limits^(38)_(28) {f(x)} \, dx = 0.57`

you can find this answer using the R statistical programming languange and the instruction pnorm(38, mean = 35, sd = 6) -pnorm(28, mean = 35, sd = 6)

d) we find this probability as

`P(41\leq X\leq 44)`=
`\int\limits^(44)_(41) {f(x)} \, dx = 0.09`

you can find this answer using the R statistical programming languange and the instruction pnorm(44, mean = 35, sd = 6) -pnorm(41, mean = 35, sd = 6)

Answer 2

a) 37.31

b) 42.68

c) 57.05% probability of getting a raw score between 28 and 38

d) 9.19% probability of getting a raw score between 41 and 44.

Explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

`\mu = 35, \sigma = 6`

a) What number represents the 65th percentile (what number separates the lower 65% of the distribution)?

This is X when Z has a pvalue of 0.65. So X when Z = 0.385.

`Z = (X - \mu)/(\sigma)`

`0.385 = (X - 35)/(6)`

`X - 35 = 6*0.385`

`X = 37.31`

b) What number represents the 90th percentile?

This is X when Z has a pvalue of 0.9. So X when Z = 1.28

`Z = (X - \mu)/(\sigma)`

`1.28 = (X - 35)/(6)`

`X - 35 = 6*1.28`

`X = 42.68`

c) What is the probability of getting a raw score between 28 and 38?

This is the pvalue of Z when X = 38 subtracted by the pvalue of Z when X = 28. So

X = 38

`Z = (X - \mu)/(\sigma)`

`Z = (38 - 35)/(6)`

`Z = 0.5`

`Z = 0.5` has a pvalue of 0.6915

X = 28

`Z = (X - \mu)/(\sigma)`

`Z = (28 - 35)/(6)`

`Z = -1.17`

`Z = -1.17` has a pvalue of 0.1210

0.6915 - 0.1210 = 0.5705

57.05% probability of getting a raw score between 28 and 38

d) What is the probability of getting a raw score between 41 and 44?

This is the pvalue of Z when X = 44 subtracted by the pvalue of Z when X = 41. So

X = 44

`Z = (X - \mu)/(\sigma)`

`Z = (44 - 35)/(6)`

`Z = 1.5`

`Z = 1.5` has a pvalue of 0.9332

X = 41

`Z = (X - \mu)/(\sigma)`

`Z = (41 - 35)/(6)`

`Z = 1`

`Z = 1` has a pvalue of 0.8413

0.9332 - 0.8413 = 0.0919

9.19% probability of getting a raw score between 41 and 44.