Question

An aptitude test has a mean score of 80 and a standard deviation of 5. The population of scores is normally distributed. What raw score corresponds to the 70th percentile?

Answer 1

Answer:

82.62

Step-by-step explanation:

Mean score (μ) = 80

Standard deviation (σ) = 5

The 70th percentile of a normal distribution has an equivalent z-score of roughly 0.525.

For any given score, X, the z-score can be determined by:

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

For z = 0.525:

$0.525=\frac{X-80}{5}\\ X=82.62$

A raw score of approximately 82.62 corresponds to the 70th percentile.

Answer 2

Answer: the raw score that corresponds to the 70th percentile is 82.625

Step-by-step explanation:

Since the population of scores in the aptitude test is normally distributed., we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = aptitude test scores.

µ = mean score

σ = standard deviation

From the information given,

µ = 80

σ = 5

We want to find the raw score that corresponds to the 70th percentile.

70th percentile = 70/100 = 0.7

Looking at the normal distribution table, the z score corresponding to 0.7 is 0.525.

Therefore,

0.525 = (x - 80)/5

5 × 0.525 = x - 80

2.625 = x - 80

x = 2.625 + 80

x = 82.625