Question

The scores on the Wechsler Adult Intelligence Scale are approximately Normal with mean = 100 and standard deviation = 15The proportion of adults with scores above 130 is closest toa) 0.001b) 0.025c) 0.050d) 0.950

Answer 1

Answer: option b is the closet

Step-by-step explanation:

Since the scores on the Wechsler Adult Intelligence Scale are approximately Normal, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = scores on the Wechsler Adult Intelligence Scale.

µ = mean score

σ = standard deviation

From the information given,

µ = 100

σ = 15

We want to find the proportion of adults with scores above 130. It is expressed as

P(x > 130) = 1 - P(x ≤ 130)

For x = 130,

z = (130 - 100)/15 = 2

Looking at the normal distribution table, the probability corresponding to the z score is 0.9773

P(x > 130) = 1 - 0.9773 = 0.023

Answer 2

Answer:

b) 0.025

Step-by-step explanation:

Problems of normally distributed samples are 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:

$\mu = 100, \sigma = 15$

The proportion of adults with scores above 130 is closest to

This proportion is 1 subtracted by the pvalue of Z when X = 130. So

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

$Z = \frac{130 - 100}{15}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772

1 - 0.9772 = 0.0228

Closest to 0.025

So the correct answer is:

b) 0.025