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adoni [48]
3 years ago
6

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

tudents scored below 71? Hint: Use the 68-95-99.7 rule.

Mathematics
2 answers:
Keith_Richards [23]3 years ago
6 0

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.

z=\frac{x-\mu}{\sigma}, where,

z=\text{z-score},

x=\text{Raw score},

\mu=\text{Mean},

\sigma=\text{Standard deviation}.

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

z=\frac{71-65}{6}

z=\frac{6}{6}=1

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.

\frac{100\%-68\%}{2}=\frac{32\%}{2}=16\%

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

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

\text{The percentage of students scored below 71}=84\%

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

tiny-mole [99]3 years ago
5 0
Standardising X=71 to find z-score

z-score= \frac{71-65}{6} =1

reading from z-table

P(Z\ \textless \ 1)=0.8413=84.13%

Students score below 71 is 84.13%

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