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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 64 and 12, respectively. If the scores are normally distributed, what percentage

In a statistics class, the mean and standard deviation of the scores obtained by students on a test were found to be 64 and 12, respectively. If the scores are normally distributed, what percentage of students scored below 69 ?

Option: 1

52.78 \%


Option: 2

96.41 \%


Option: 3

66.15 \%


Option: 4

86.66 \%


Answers (1)

best_answer

To determine the percentage of students who scored below 69 , 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, 69),
\mu is the mean (64), and
\sigma is the standard deviation (12).

Plugging in the values:

\mathrm{ \begin{aligned} & z=(69-64) / 12 \\ & z=5 / 12 \\ & z \approx 0.4167 \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 69 .

Looking up the z-score of 0.4167 in the z-table, we find that the area to the left is approximately 0.6615 .

So, approximately 66.15% of the students scored below 69 on the test.

 

Posted by

Gautam harsolia

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