Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
1. Section 10.1
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Exercise 67 Page 520

Practice makes perfect
a

Let's draw a normal distribution of the average height of adult women. We will also include the interval of women that are under 4 feet and 11 inches tall. To do this, we must first rewrite the given height in inches only. We can convert from feet to inches if we multiply by 12.

4(12)+11= 59 inches Let's now draw the normal distribution. We will also mark the interval of heights less than or equal to 59 inches.

To determine the percentage of women that represent this interval, we can use the normal distribution function on a graphing calculator. On the calculator push 2nd, then VARS, and pick the second option.

Window with a graph

Now, we need to type the parameters of the distribution. Recall that the upper limit is 66. For the lower limit we will put some arbitrary very low number to make sure we cover nearly everything in the left tail of the normal distribution. We are also given that μ= 63.8 and σ = 2.7.

Window with a graph

Having entered all of the values, push ENTER until we get a result.

Window with a graph

As we can see, about 3.77 % of women are less than 4 feet 11 inches tall.

b

If we assume that half the senior students are girls, we get the following number of girls in the class.

324/2=162 From Part B, we identified that about 3.77 % of all girls are below 4 feet 11 inches tall. Therefore, if we multiply the number of girls by this percentage, we can determine how many girls are below this height. 0.0377(162)≈ 6 About 6 girls are below 4 ft 11 inches.
c

To determine the number of girls that are expected to be above 6 feet, we will first convert it to inches.

6(12)= 72inches This time the lower limit is 72 inches, and for the upper limit we put something very high to make sure we cover nearly everything to the right of the lower limit. Recall that μ= 63.8 and σ = 2.7.

Window with a graph
Window with a graph

About 0.12 % of girls are above 6 feet. Similarly as in Part B, let's calculate the expected number of senior girls taller than 6ft. 0.00119(162)≈ 0 Therefore, we expect 0 senior girls to be taller than 6feet.