Houghton Mifflin Harcourt Algebra 1, 2015
HM
Houghton Mifflin Harcourt Algebra 1, 2015 View details
4. Normal Distributions
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Exercise 1 Page 342

Find the difference between the mean score and the given score in terms of the number of standard deviations.

97.5 %

Practice makes perfect

Let's begin by reviewing the properties of a normal curve.

  • About 68 % of the data values fall within 1 standard deviation of the mean.
  • About 95 % of the data values fall within 2 standard deviations of the mean.
  • About 99.7 % of the data values fall within 3 standard deviations of the mean.

We can look at a diagram to better visualize these properties.

empirical rule button graph
Now, we are told that the scores on a test are normally distributed with a mean of 74 and a standard deviation of 8. Using this information we want to find the percentage of juniors whose score is no more than 90. x≤ 90 First, let's find the difference between 90 and the mean 74. 90- 74= 16 Next, we will divide the difference by the standard deviation 8. 16/8= 2 Therefore, 90 is 2 standard deviations above the mean. Let's draw a normal curve for the given mean and standard deviation. To find the percentage, we will shade the area that represents values no more than 2 standard deviations above the mean.

Finally, we can find the percentage of juniors by adding the percentages of the shaded areas.

p=0.15 %+2.35 %+13.5 %+34 %+34 %+13.5 %
p=97.5 %

The percentage of juniors who scored no more than 90 is 97.5 %.