High School

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------------------------------------------------ A class's exam scores are normally distributed. If the average score is 65 and the standard deviation is 6, what percentage of students scored below 71?

Hint: Use the 68-95-99.7 rule.

Answer :

Standardising X=71 to find z-score

[tex]z-score= \frac{71-65}{6} =1[/tex]

reading from z-table

[tex]P(Z\ \textless \ 1)=0.8413=84.13%[/tex]

Students score below 71 is 84.13%

Answer:

84%.

Step-by-step explanation:

We have been given that a A class's exam scores are normally distributed. The average score is 65 and the standard deviation is 6.

We will use the z-score formula to find the z-score corresponding to raw score of 71.

[tex]z=\frac{x-\mu}{\sigma}[/tex], where,

[tex]z=\text{z-score}[/tex],

[tex]x=\text{Raw score}[/tex],

[tex]\mu=\text{Mean}[/tex],

[tex]\sigma=\text{Standard deviation}[/tex].

Upon substituting our given values in z-score formula we will get,

[tex]z=\frac{71-65}{6}[/tex]

[tex]z=\frac{6}{6}=1[/tex]

Since we know that 68-95-99.7 rule states that approximately 68%, 95% and 99.7% of data lies within one, two and three standard deviation of mean respectively.

Since 68% of data lies within one standard deviation of mean. Now we subtract 68% from 100% and divide the result by 2.

[tex]\frac{100\%-68\%}{2}=\frac{32\%}{2}=16\%[/tex]

Now we will add 16% to 68% to get the percentage of students that scored below 71.

[tex]\text{The percentage of students scored below 71}=68\%+16\%[/tex]

[tex]\text{The percentage of students scored below 71}=84\%[/tex]

Therefore, approximately 84% of the students scored below 71.