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  • In a statistics class, the mean and standard deviation of the scores obtained by students on a test were found to be 59 and 16 , respectively. If the scores are normally distributed, what percentag

In a statistics class, the mean and standard deviation of the scores obtained by students on a test were found to be 59 and 16 , respectively. If the scores are normally distributed, what percentage of students scored below 63 ?

Option: 1

76.36 \%


Option: 2

66.52 \%


Option: 3

59.87 \%


Option: 4

77.63 \%


Answers (1)

best_answer

To determine the percentage of students who scored below 63 , we need to calculate the z-score corresponding to that value and then find the area under the normal curve to the left of that z-score.

The z-score is calculated using the formula:

\mathrm{ z=(x-\mu) / \sigma }
where:
\mathrm{x}  is the value we want to find the percentage below (in this case, 63),
\mu is the mean (59), and
\sigma is the standard deviation (16).

Plugging in the values:

\mathrm{ \begin{aligned} & z=(63-59) / 16 \\ & z=4 / 16 \\ & z=0.25 \end{aligned} }

Now we need to find the area to the left of this z-score in the standard normal distribution table (also known as the z-table) or by using a statistical software. The area represents the percentage of students who scored below 63 .

Looking up the z-score of 0.25 in the z-table, we find that the area to the left is approximately 0.5987 .

So, approximately 59.87 % of the students scored below 63 on the test. 

Posted by

Ajit Kumar Dubey

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