Question

Suppose that scores on a particular test are normally distributed with a mean of 140 and a standard deviation of 20. What is the minimum score needed to be in the top 5% of the scores on the test? Carry your intermediate computations to at least four decimal places, and round your answer to at least one decimal place.

Answer 1

Answer:

The minimum score needed to be in the top 5% of the scores on the test is 172.9.

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:

$\mu = 140, \sigma = 20$

What is the minimum score needed to be in the top 5% of the scores on the test?

The 100-5 = 95th percentile, which is the value of X when Z has a pvalue of 0.95. So it is X when Z = 1.645.

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

$1.645 = \frac{X - 140}{20}$

$X - 140 = 20*1.645$

$X = 172.9$

The minimum score needed to be in the top 5% of the scores on the test is 172.9.

Answer 2

Answer: the minimum score needed to be in the top 5% of the scores on the test is 173

Step-by-step explanation:

Suppose that scores on the particular test are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = scores on the test.

µ = mean score

σ = standard deviation

From the information given,

µ = 140

σ = 20

The probability value for the minimum score needed to be in the top 5% of the scores on the test would be (1 - 5/100) = (1 - 0.05) = 0.95

Looking at the normal distribution table, the z score corresponding to the probability value is 1.65

Therefore,

1.65 = (x - 140)/20

Cross multiplying by 20, it becomes

1.65 × 20 = x - 140

33 = x - 140

x = 33 + 140

x = 173