Practice 2 Solutions

Name the area for the standard normal curve as a proportion and a percent.

$z = 3.02$

$P (z < 3.02) = 0.9987 = 99.87 %$

$z = - 2.34$

$P (z < - 2.34) = 0.0096 = 0.96 %$

Determine the z-score given the proportion.

0.3970

$z = - 0.26$

0.3781

$z = - 0.31$

For problems 5–8, calculate the z-score. Then determine the percentage or raw data score.

Mr. Bartz’s class recorded their nightly reading times in minutes over a two-week period. He calculated the class average to be 22 minutes with a standard deviation of 3 minutes.

What percentage of the class read less than 15 minutes per night?

$\begin{array}{rcl} z & = & \frac{X - μ}{σ} \\ z & = & \frac{15 - 22}{3} \\ z & = & - 2.33 \\ P (X < 15) & = & 0.0099 \end{array}$

In Mr. Bartz’s class, 0.99% of students read less than 15 minutes per night.

What percentage of Mr. Bartz’s students read more than 25 minutes per night?

$\begin{array}{rcl} z & = & \frac{X - μ}{σ} \\ z & = & \frac{25 - 22}{3} \\ z & = & 1.000 \\ P (X < 25) & = & 0.8413 \\ 1 - 0.8413 & = & 0.1587 \end{array}$

About 16% of students read more than 25 minutes per night.

Note

Q: If the z-table represents the area below the z-score, how do you find data above the z-score?

A: Subtract the area from one

Q: From the Empirical rule, what is the percentage of data greater than one standard deviation above the mean?

A: $13.5 % + 2.35 % + 0.015 % = 15.865 %$

Reference the image in the notes if needed.

What percentage of students read between 17 and 20 minutes per night?

$z = \frac{X - μ}{σ} \begin{array}{ll} X = 17 & X = 20 \\ z = \frac{17 - 22}{3} & z = \frac{20 - 22}{3} \\ z = - 1.67 & z = - 0.67 \\ P (X < 17) = 0.0475 & P (X < 20) = 0.2514 \\ 0.2514 - 0.0475 = 0.2039 \end{array}$

About 20% of students read between 17 and 20 minutes per night.

How long, to the nearest minute, did 66% of Mr. Bartz’s class read?

$\begin{array}{rcl} 0.66 & \approx & 0.6591 \\ z & = & 0.41 \\ 0.41 & = & \frac{X - 22}{3} \\ 1.23 & = & X - 22 \\ X & = & 23.23 \end{array}$

About 66% of the class read for 23 minutes or less.

For problems 9–12, use the scenario.

Ms. Troglin’s botany class studied the growth of Flower B plants in centimeters (cm) over thirty days. On day zero, students planted seeds. Then on day thirty, students measured the height of each Flower B plant. Using their measurements, Ms. Troglin’s class calculated that the average height of the plants was 26.4 cm with a 1.2 cm standard deviation.

What height does Ms. Troglin’s class expect 40% of all Flower B plants to be?

$\begin{array}{rcl} z & = & - 0.25 \\ - 0.25 & = & \frac{X - 26.4}{1.2} \\ - 0.3 & = & X - 26.4 \\ X & = & 26.1 \end{array}$

The class predicts that 40% of the plants are 26.1 cm or less in height.

Determine the height of the top 5.05% of the plants.

$\begin{array}{rcl} 1 - 0.0505 & = & 0.9495 \\ z & = & 1.64 \\ 1.64 & = & \frac{X - 26.4}{1.2} \\ 1.968 & = & X - 26.4 \\ X & = & 28.37 \end{array}$

The top 5.05% of plant heights were 28.37 cm or more.

What percentage of plants were less than 23 centimeters?

$\begin{array}{rcl} X & = & 23 \\ z & = & \frac{23 - 26.4}{1.2} \\ z & = & - 2.83 \end{array}$

$P (height < 23) = 0.23 %$

What percentage of plants were between 25 and 27 centimeters?

$z = \frac{X - μ}{σ} \begin{array}{ll} X = 25 & X = 27 \\ z = \frac{25 - 26.4}{1.2} & z = \frac{27 - 26.4}{1.2} \\ z = - 1.17 & z = 0.5 \\ P (X < 25) = 0.1210 & P (X < 27) = 0.6915 \\ 0.6915 - 0.1210 = 0.5705 \end{array}$

57.05% of plants were between 25 cm and 27 cm in height.

For problems 13–14, determine the missing value in the formula $z = \frac{X - μ}{σ} .$

$P = 0.1093, X = 13.2, σ = 7.51$

$\begin{array}{rcl} z & = & - 1.23 \\ - 1.23 & = & \frac{13.2 - μ}{7.51} \\ - 9.24 & = & 13.2 - μ \\ μ & = & 22.44 \end{array}$

$μ = 22.44$

$z = 0.72, μ = 18, σ = 4.35$

$\begin{array}{rcl} 0.72 & = & \frac{X - 18}{4.35} \\ 3.13 & = & X - 18 \end{array}$

$X = 21.13$

For problems 15–16, use the scenario.

A fisheries biologist calculated the average length of a lake fish population to be 10.12 centimeters and the standard deviation to be 1.57 centimeters. Then, one year later, a sample of the population was measured.

Fish Length (cm)
8.0	8.4	9.5
9.4	7.3	7.4
6.2	8.5
10.1	10.0

Complete the table to determine if the sample group of fish represents a normal data set.

z	–2	–1	0	1	2
Area < z	0.02	0.16	0.5	0.84	0.98
Predicted: 10z	$0.2 \approx 0$	$1.6 \approx 2$	6	$8.4 \approx 8$	$9.8 \approx 10$
$X < 𝜇 \pm 2 𝜎$	$X < 6.98$	$X < 8.55$	$X < 10.12$	$X < 11.69$	$X < 13.26$
Actual Count	1	6	all	all	all

Is the sample normally distributed? Explain.

Sample: The sample is not normally distributed because all of the values are below the mean.

Note

The population mean is not representative of the sample.