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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 69 and 13 , 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 69 and 13 , respectively. If the scores are normally distributed, what percentage of students scored below 74 ?

Option: 1

86.24 \%


Option: 2

66.14 \%


Option: 3

64.90 \%


Option: 4

63.94 \%


Answers (1)

best_answer

To determine the percentage of students who scored below 74 , 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, 74),
\mu is the mean (69), and
\sigma is the standard deviation (13).

Plugging in the values:

\mathrm{ \begin{aligned} & z=(74-69) / 13 \\ & z=5 / 13 \\ & z \approx 0.3846 \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 74 .

Looking up the z-score of 0.3846 in the z-table, we find that the area to the left is approximately 0.6490 .

So, approximately 64.90 % of the students scored below 74 on the test.

Posted by

Ritika Harsh

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