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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 58 and 14, respectively. If the scores are normally d

In a statistics class, the mean and standard deviation of the scores obtained by students on a test were found to be 58 and 14, respectively. If the scores are normally distributed, what percentage of students scored below 63?

 

Option: 1

\quad 52.99 \%


Option: 2

96.54 \%


Option: 3

64.49 \%


Option: 4

86.33 \%


Answers (1)

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:

x  is the value we want to find the percentage below (in this case, 63),
\mu is the mean (58), and
\sigma  is the standard deviation (14).

Plugging in the values:

\mathrm{ \begin{aligned} & z=(63-58) / 14 \\ & z=5 / 14 \\ & z \approx 0.3571 \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 0.3571 in the z-table, we find that the area to the left is approximately 0.6449 .

So, approximately 64.49% of the students scored below 63 on the test.

 

Posted by

Sumit Saini

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