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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 50 and 20 , respectively. If the scores are normally distributed, what percentage of students scored below 65 ?

Option: 1

12.35 \%


Option: 2

88.36 \%


Option: 3

96.32 \%


Option: 4

77.34 \%


Answers (1)

best_answer

To determine the percentage of students who scored below 65 , 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, 65),
\mu is the mean (50), and
\sigma is the standard deviation (20).

Plugging in the values:

\mathrm{ \begin{aligned} & z=(65-50) / 20 \\ & z=15 / 20 \\ & z=0.75 \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 65 .

Looking up the z-score of 0.75 in the z-table, we find that the area to the left is approximately 0.7734 .

So, approximately 77.34% of the students scored below 65 on the test.

 

Posted by

Devendra Khairwa

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