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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 67 and 11, respectively. If the scores are normally d

In a statistics class, the mean and standard deviation of the scores obtained by students on a test were found to be 67 and 11, respectively. If the scores are normally distributed, what percentage of students scored below 72?

 

Option: 1

52.85 \%


Option: 2

67.36 \%


Option: 3

77.36 \%


Option: 4

99.65 \%


Answers (1)

best_answer

To determine the percentage of students who scored below 72 , we can use the properties of the normal distribution.

First, we need to calculate the Z-score for a score of 72 . The Z-score represents the number of standard deviations a particular value is from the mean.

The formula to calculate the Z-score is:

\mathrm{ Z=(X-\mu) / \sigma }
Where:

\mathrm{ \begin{aligned} Z & =Z \text {-score } \\ X & =\text { Score } \\ \mu & =\text { Mean } \\ \sigma & =\text { Standard deviation } \end{aligned} }

Plugging in the values:

\mathrm{ \begin{aligned} & Z=(72-67) / 11 \\ & Z=5 / 11 \\ & Z \approx 0.4545 \end{aligned} }

Next, we need to find the cumulative probability associated with the Z-score of 0.4545 . This represents the percentage of values below 72 .

Using a standard normal distribution table or a calculator, we can find that the cumulative probability associated with a Z-score of 0.4545 is approximately 0.6736 .

To convert this probability to a percentage, we multiply it by 100 :

\mathrm{ \begin{aligned} \text { Percentage } & =0.6736 \times 100 \\ \text { Percentage } & \approx 67.36 \% \end{aligned} }
Therefore, approximately 67.36 \% of students scored below 72 on the test.

Posted by

Ajit Kumar Dubey

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