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The distribution of scores on a recent test closely followed a Normal Distribution with a mean of 22 points and a standard deviation of 2 points. For this question, DO NOT apply the standard deviation rule.
(a) What proportion of the students scored at least 26 points on this test, rounded to five decimal places?
.02275
(b) What is the 71 percentile of the distribution of test scores, rounded to three decimal places?
24.073

Answer :

The proportion of the students for the given condition is 0.0228 and the 71 percentile of the normal distribution of the test score is equal to 23.106.

As given in the question,

Mean of the normal distribution 'μ' = 22 points

Standard deviation of the normal distribution 'σ' = 2 points

Z score is calculated using :

z = (x - μ)/σ

Here, x represents the raw score

a. Condition: Students score at least 26 points

x > 26

⇒ (x - μ) / σ > ( 26 - 22 ) / 2

⇒ (x - μ) / σ > 4 /2

⇒ (x - μ) / σ > 2

⇒ z > 2

Proportion of the students scored z > 2

P(z > 2) = 1 - P(z < 2)

             = 1 -  0.9772

             = 0.0228

b. 71 percentile has a z score of 0.553

0.553 = (x - 22)/2

⇒x - 22 = 1.106

⇒ x = 23.106

Therefore, the proportion of the students scored at least 26 points are 0.0228 and 71 percentile of the normal distribution of the test score is equal to 23.106.

Learn more about normal distribution here

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