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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 74 and 18 , respectively. If the scores are normally distributed, what percentage of students scored below 80 ?

Option: 1

32.85 \%


Option: 2

96.74 \%


Option: 3

62.93 \%


Option: 4

64.36 \%


Answers (1)

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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 (74), and

\sigma is the standard deviation (18).

Plugging in the values:

\mathrm{ \begin{aligned} & z=(80-74) / 18 \\ & z=6 / 18 \\ & z=1 / 3 \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 1 / 3 in the z-table, we find that the area to the left is approximately 0.6293 .

So, approximately 62.93 % of the students scored below 80 on the test.

Posted by

Kuldeep Maurya

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