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Schach [20]
3 years ago
9

Scores on an exam follow an approximately bell shaped distribution with a mean of 76.4 and a standard deviation of 6.1 points. A

pproximately, what percentage of the data is between 64.2 points and 88.6 points?
a) 68%
b) 95%
c) 99.7%
d) 50%
Mathematics
1 answer:
Crazy boy [7]3 years ago
6 0

Answer:

b) 95%

Step-by-step explanation:

We have been given that scores on an approximately bell shaped distribution with a mean of 76.4 and a standard deviation of 6.1 points. We are asked to find the percentage of the data that is between 64.2 points and 88.6 points.

First of all, we will find z-scores of each data point as:

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

z=\frac{64.2-76.4}{6.1}

z=\frac{-12.2}{6.1}

z=-2

Let us find z-score corresponding to normal score 88.6.

z=\frac{88.6-76.4}{6.1}

z=\frac{12.2}{6.1}

z=2

To find the percentage of the data is between 64.2 points and 88.6 points, we need to find area under a normal distribution curve that lie within two standard deviation of mean.

The empirical rule of normal distribution states that approximately 95% of data points fall within two standard deviation of mean, therefore, option 'b' is the correct choice.

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WARRIOR [948]

It should be noted that No. the value of the 9 in the hundred place is not ten times the value of 3 in the tens place.

<h3>How to illustrate the information?</h3>

It should be noted that the value of 3 in 1934 is 30 and the value of 9 is 900.

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Therefore the value of 9 will be:

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Therefore, the the value of the 9 in the hundred place is 30 times the value of 3 in the tens place.

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Step-by-step explanation:

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