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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 62 and 9, respectively. If the scores are normally di

In a statistics class, the mean and standard deviation of the scores obtained by students on a test were found to be 62 and 9, respectively. If the scores are normally distributed, what percentage of students scored below 70?

 

Option: 1

81.21 \%


Option: 2

77.36 \%


Option: 3

45.36 \%


Option: 4

65.98 \%


Answers (1)

best_answer

To determine the percentage of students who scored below 70 , 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, 70),
\mu is the mean (62), and
\sigma is the standard deviation (9).

Plugging in the values:

\mathrm{ \begin{aligned} & z=(70-62) / 9 \\ & z=8 / 9 \\ & z \approx 0.8889 \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 70 .

Looking up the z-score of 0.8889 in the z-table, we find that the area to the left is approximately 0.8121 .

So, approximately 81.21% of the students scored below 70 on the test.

 

Posted by

manish painkra

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