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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 70 and 15 , respectively. If the scores are normally distributed, what percentage of students scored below 80 ?

Option: 1

74.75 \%


Option: 2

58.47 \%


Option: 3

36.14 \%


Option: 4

96.25 \%


Answers (1)

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:

x is the value we want to find the percentage below (in this case, 80 ),
\mu is the mean (70), and
\sigma is the standard deviation (15).

Plugging in the values:

\mathrm{ \begin{aligned} & z=(80-70) / 15 \\ & z=10 / 15 \\ & z=0.6667 \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 0.6667 in the z-table, we find that the area to the left is approximately 0.7475 .

So, approximately 74.75% of the students scored below 80 on the test.

 

Posted by

Ramraj Saini

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