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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 72 and 13 , respectively. If the scores are normally distributed, what percentage of students scored below 75 ?

Option: 1

22.36 \%


Option: 2

96.45 \%


Option: 3

72.96 \%


Option: 4

58.90 \%


Answers (1)

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To determine the percentage of students who scored below 75 , 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:
x is the value we want to find the percentage below (in this case, 75),

\mu is the mean (72), and

\sigma is the standard deviation (13).

Plugging in the values:

\mathrm{ \begin{aligned} & z=(75-72) / 13 \\ & z=3 / 13 \\ & z \approx 0.2308 \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 75 .

Looking up the z-score of 0.2308 in the z-table, we find that the area to the left is approximately 0.5890 .

So, approximately 58.90 % of the students scored below 75 on the test. 

Posted by

Divya Prakash Singh

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