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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 64 and 14, 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 64 and 14, respectively. If the scores are normally distributed, what percentage of students scored below 69?

 

Option: 1

22.13%


Option: 2

64.18%


Option: 3

54.36%


Option: 4

83.56%


Answers (1)

best_answer

To determine the percentage of students who scored below 69, we can use the properties of a normal distribution. Given that the mean (\mu) is 64 and the standard deviation (\sigma) is 14, we can calculate the z-score corresponding to a score of 69 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:

\mathrm{x} is the individual score
\mathrm{\mu} is the mean
\mathrm{\sigma } is the standard deviation

In this case, we want to find the z-score for \mathrm{x= 69}:

\begin{aligned} & z=(69-64) / 14 \\ & z=0.3571 \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 0.3571 is approximately 64.18%.

Therefore, approximately 64.18% of students scored below 69.

 

Posted by

HARSH KANKARIA

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