Question

On a nationwide test taken by high school students, the mean score was 53 and the standard deviation was 12. The scores were normally distributed. Complete the following statements.
Approximately 99.7% of the students scored between ? and ?.
Approximately ?% of the students scored between 41 and 65.

Answer 1

99.7% of the students scored between (44.67 , 56.33) i.e.

between 44 and 56.

On Subtracting 1 from your sample size, we get

100 – 1 = 99.

On Subtracting confidence level from 1, and then divide by two.

(1 – .997) / 2 = 0.0015

Now, df = 99 and α = 0.0015

from the table at df = 99 we got 2.262.

Divide your sample standard deviation by the square root of your sample size.

28 / √(100) = 2.8

Now, 2.262 × 2.8 = 6.33

So, the confidence interval be,

(100 - 6.33 , 100 + 6.33) = (44.67 , 56.33)

Hence, 99.7% of the students scored between (44.67 , 56.33) i.e.

between 44 and 56.

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