Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
10. Normal Distributions
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Exercise 24 Page 744

To start solving, find the difference between the mean of the data set, and the lower and upper bounds of the given interval.

99.7 %

Practice makes perfect
We know that the set of data is normally distributed with a mean of 100 and a standard deviation of 10. We will find the probability of choosing a value that is between 70 and 130. 70 ≤ x≤ 130 First, let's find the difference between 70 and the mean 100. 70- 100= - 30

Now, let's find the difference, between 130 and the mean 100. 130- 100= 30 Then, we will divide the differences by the standard deviation 10. - 30/10=- 3and30/10=3 Thus, 70 is 3 standard deviations below the mean and 130 is 3 standard deviations above the mean. To find the probability, we will shade the percent of data which is at most 3 standard deviations below the mean and no more than 3 standard deviations above the mean.

Finally, we can find the percent of data by adding the percents of the shaded areas.
p=2.35 %+13.5 %+34 %+34 %+13.5 %+2.35 %
p=99.7 %
The probability of choosing a random value between 70 and 130 is equal to 99.7 %.