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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 71 and 15, 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 71 and 15, respectively. If the scores are normally distributed, what percentage of students scored below 73?

 

Option: 1

55.59%


Option: 2

66.12%


Option: 3

22.12%


Option: 4

63.86


Answers (1)

best_answer

To determine the percentage of students who scored below 73, we can use the properties of a normal distribution. Given that the mean (\mu) is 71 and the standard deviation (\sigma) is 15, we can calculate the z-score corresponding to a score of 73 and then find the percentage of scores below that z-score.

The z-score is calculated using the formula:

\mathrm{z=(x-\mu) / \sigma}
Where:
x  is the individual score
\muis the mean
\sigma s the standard deviation

In this case, we want to find the z-score for x= 73:
\begin{aligned} & z=(73-71) / 15 \\ & z=0.1333 \end{aligned}

Once we have the z-score, we can look up the corresponding percentage in the standard normal distribution table or use a calculator that provides this functionality. From the table, we find that the percentage of scores below a z-score of 0.1333 is approximately 55.59%.

Therefore, approximately 55.59% of students scored below 73.

 

Posted by

Suraj Bhandari

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