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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 76 and 7 , 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 76 and 7 , respectively. If the scores are normally distributed, what percentage of students scored below 80 ?

Option: 1

52.12 \%


Option: 2

96 \%


Option: 3

78 \%


Option: 4

71.32 \%


Answers (1)

best_answer

To determine the percentage of students who scored below 80 , 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:
\mathrm{x} is the value we want to find the percentage below (in this case, 80 ),
\mu is the mean (76), and
\sigma is the standard deviation (7).

Plugging in the values:

\mathrm{ \begin{aligned} & z=(80-76) / 7 \\ & z=4 / 7 \\ & z \approx 0.5714 \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 80 .

Looking up the z-score of 0.5714 in the z-table, we find that the area to the left is approximately 0.7132 .

So, approximately 71.32% of the students scored below 80 on the test.

 

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Pankaj

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