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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 58 and 16 , respectively. If the scores are normally distributed, what percentage of students scored below 63 ?

Option: 1

62.21 \%


Option: 2

34.15 \%


Option: 3

72.78 \%


Option: 4

77.96 \%


Answers (1)

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To determine the percentage of students who scored below 63 , we can use the properties of the normal distribution.

First, we need to calculate the Z-score for a score of 63. 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=(63-58) / 16 \\ & Z=5 / 16 \\ & Z \approx 0.3125 \end{aligned} }

Next, we need to find the cumulative probability associated with the Z-score of 0.3125 . This represents the percentage of values below 63 .

Using a standard normal distribution table or a calculator, we can find that the cumulative probability associated with a Z-score of 0.3125 is approximately 0.6221 .
To convert this probability to a percentage, we multiply it by 100 :

\mathrm{ \begin{aligned} & \text { Percentage }=0.6221 \times 100 \\ & \text { Percentage } \approx 62.21 \% \end{aligned} }

Therefore, approximately 62.21 \% of students scored below 63 on the test.

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