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

Answer:

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

$f(x) = \frac{1}{\sqrt{2\pi}\sigma}\exp[-\frac{(x-\mu)^{2}}{2\sigma^{2}}]$

in this case we have

$f(x) = \frac{1}{\sqrt{2\pi}6}\exp[-\frac{(x-35)^{2}}{2(6^{2})}]$

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

Answer:

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.

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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.

$Z = \frac{X - \mu}{\sigma}$

$0.385 = \frac{X - 35}{6}$

$X - 35 = 6*0.385$

$X = 37.31$

b) What number represents the 90th percentile?

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

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 35}{6}$

$X - 35 = 6*1.28$

$X = 42.68$

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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{38 - 35}{6}$

$Z = 0.5$

$Z = 0.5$ has a pvalue of 0.6915

X = 28

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{28 - 35}{6}$

$Z = -1.17$

$Z = -1.17$ 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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{44 - 35}{6}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

X = 41

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{41 - 35}{6}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

0.9332 - 0.8413 = 0.0919

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