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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 73 and 10 , respectively. If the scores are normally distributed, what percentage of students scored below 78 ?

Option: 1

54.78 \%


Option: 2

66.32 \%


Option: 3

78.36 \%


Option: 4

69.15 \%


Answers (1)

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To determine the percentage of students who scored below 78 , we can use the properties of the normal distribution.

First, we need to calculate the Z-score for a score of 78. 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:

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=(78-73) / 10 \\ & Z=5 / 10 \\ & Z=0.5 \end{aligned} }

Next, we need to find the cumulative probability associated with the Z-score of 0.5 . This represents the percentage of values below 78 .

Using a standard normal distribution table or a calculator, we can find that the cumulative probability associated with a Z-score of 0.5 is approximately 0.6915 .

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

\mathrm{\begin{aligned} & \text { Percentage }=0.6915 \times 100 \\ & \text { Percentage } \approx 69.15 \% \end{aligned}}

Therefore, approximately 69.15 \% of students scored below 78 on the test.

Posted by

Ajit Kumar Dubey

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