The Z Score

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Question 1 of 4

1. Question
If a set of scores have a mean of $11$ $11$ and a standard deviation of $0.9$ $0.9$ , sketch the normal curve then find the z-score for $12$ $12$ .
- 1. $2.12$ $2.12$
- 2. $1.52$ $1.52$
- 3. $1.8$ $1.8$
- 4. $1.11$ $1.11$
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$Z$ $Z$ score

$Z = \frac{x - \bar{x}}{S}$ $Z=(x-barx)/S$

Given Values

Mean $\bar{X} = 11$ $barX= 11$

Standard Deviation $= 0.9$ $=0.9$

First, complete the labels of the bell curve by using the mean and standard deviation.

For example, start with the mean, $11$ $11$ . Then add and subtract $0.9$ $0.9$ to get the values $1$ $1$ standard deviation above and below the mean.

Keep adding and subtracting the standard deviation until the labels are completed.

Add the z-score values.

Input the values to the z-score formula to solve for the z-score.

$Z$ $Z$ $=$ $=$ $\frac{x - \bar{x}}{S}$ $(x-barx)/S$

$=$ $=$ $\frac{12 - 11}{0.9}$ $(12-11)/0.9$ Substitute values

$=$ $=$ $1.11$ $1.11$ Simplify

$1.11$ $1.11$
Question 2 of 4

2. Question
Given that the mean is $78.4$ $78.4$ and the standard deviation is $3.2$ $3.2$ , find the z-scores for $68.2$ $68.2$ and $82.4$ $82.4$ .
- $68.2$ $68.2$ z-score: (-3.19, -3.1875)
  
  $82.4$ $82.4$ z-score: (1.25)
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$Z$ $Z$ score

$Z = \frac{x - \bar{x}}{S}$ $Z=(x-barx)/S$

Given Values

Mean $\bar{X} = 78.4$ $barX= 78.4$

Standard Deviation $= 3.2$ $=3.2$

First, complete the labels of the bell curve by using the mean and standard deviation.

For example, start with the mean, $78.4$ $78.4$ . Then add and subtract $3.2$ $3.2$ to get the values $1$ $1$ standard deviation above and below the mean.

Keep adding and subtracting the standard deviation until the labels are completed.

Add the z-score values.

Input the values to the z-score formula to solve for the z-score of $68.2$ $68.2$ .

$Z$ $Z$ $=$ $=$ $\frac{x - \bar{x}}{S}$ $(x-barx)/S$

$=$ $=$ $\frac{68.2 - 78.4}{3.2}$ $(68.2-78.4)/3.2$ Substitute values

$=$ $=$ $\frac{- 10.2}{3.2}$ $(-10.2)/3.2$ Simplify

$=$ $=$ $- 3.19$ $-3.19$

Do the same to solve for the z-score of $82.4$ $82.4$ .

$Z$ $Z$ $=$ $=$ $\frac{x - \bar{x}}{S}$ $(x-barx)/S$

$=$ $=$ $\frac{82.4 - 78.4}{3.2}$ $(82.4-78.4)/3.2$ Substitute values

$=$ $=$ $\frac{4}{3.2}$ $4/3.2$ Simplify

$=$ $=$ $1.25$ $1.25$

$68.2$ $68.2$ z-score: $- 3.19$ $-3.19$

$82.4$ $82.4$ z-score: $1.25$ $1.25$
Question 3 of 4

3. Question
Jack had a test for both Math and English. Given the following information, in which of the two subjects did he perform better?
- 1. English
- 2. Math
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$Z$ $Z$ score

$Z = \frac{x - \bar{x}}{S}$ $Z=(x-barx)/S$

First, find the z-score for Math.

$Z$ $Z$ $=$ $=$ $\frac{x - \bar{x}}{S}$ $(x-barx)/S$

$=$ $=$ $\frac{86 - 62}{13}$ $(86-62)/13$ Substitute values

$=$ $=$ $\frac{24}{13}$ $(24)/13$

$=$ $=$ $1.85$ $1.85$

First, find the z-score for English.

$Z$ $Z$ $=$ $=$ $\frac{x - \bar{x}}{S}$ $(x-barx)/S$

$=$ $=$ $\frac{86 - 73}{6}$ $(86-73)/6$ Substitute values

$=$ $=$ $\frac{13}{6}$ $13/6$

$=$ $=$ $2.17$ $2.17$

In this problem, the better score is determined by how further away it is from the right of the mean.
Visualizing the z-scores, we have:

Therefore, Jack performed better in English

English

Question 4 of 4

Women’s heights in a survey had a mean of

166.8

$166.8$ cm, a standard deviation of

3.1

$3.1$ cm, and is normally distributed.

(a) Where do most heights almost certainly lie?

(b) Find the z-score for the height of

173.5

$173.5$ cm.

- 1.8

$-1.8$ .

(a) (157.5)cm to (176.1)cm

(b) z-score: (2.16)

(c) height: (161.22)cm

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$Z$ $Z$ score

Z = \frac{x - \bar{x}}{S}

$Z=(x-barx)/S$

Given Values

Mean $\bar{X} = 166.8$ $barX= 166.8$ $c m$ $cm$

Standard Deviation $= 3.1$ $=3.1$ $c m$ $cm$

First, complete the labels of the bell curve by using the mean and standard deviation.

For example, start with the mean, $166.8$ $166.8$ $c m$ $cm$ . Then add and subtract $3.1$ $3.1$ $c m$ $cm$ to get the values

1

$1$ standard deviation above and below the mean.

Keep adding and subtracting the standard deviation until the labels are completed.

Add the z-score values.

Solution for (a)

99.7 %

$99.7%$ of scores lie within

3

$3$ standard deviations away from the mean.

Hence, almost all heights measure $157.5$ $157.5$ $c m$ $cm$ to $176.1$ $176.1$ $c m$ $cm$

For part (b), input the values to the z-score formula to solve for the z-score of the height,

173.5

$173.5$

c m

$cm$ .

$Z$ $Z$	$=$ $=$	$\frac{x - \bar{x}}{S}$ $(x-barx)/S$

	$=$ $=$	$\frac{173.5 - 166.8}{3.1}$ $(173.5-166.8)/3.1$	Substitute values

	$=$ $=$	$\frac{6.7}{3.1}$ $6.7/3.1$	Simplify

	$=$ $=$	$2.16$ $2.16$

The actual height of

173.5

$173.5$ cm is equivalent to a z-score of $2.16$ $2.16$ .

For part (c), input the known values into the z-score formula to solve for the corresponding height of the z-score,

- 1.8

$-1.8$ .

$Z$ $Z$	$=$ $=$	$\frac{x - \bar{x}}{S}$ $(x-barx)/S$

$- 1.8$ $-1.8$	$=$ $=$	$\frac{x - 166.8}{3.1}$ $(x-166.8)/3.1$	Substitute values

$- 1.8$ $-1.8$ $\times 3.1$ $xx3.1$	$=$ $=$	$\frac{x - 166.8}{3.1}$ $(x-166.8)/3.1$ $\times 3.1$ $xx3.1$	Multiply both sides by $3.1$ $3.1$

$- 5.58$ $-5.58$	$=$ $=$	$x - 166.8$ $x-166.8$	Simplify

$- 5.58$ $-5.58$ $+ 166.8$ $+166.8$	$=$ $=$	$x - 166.8$ $x-166.8$ $+ 166.8$ $+166.8$	Add $166.8$ $166.8$ to both sides
$161.22$ $161.22$	$=$ $=$	$x$ $x$
$x$ $x$	$=$ $=$	$161.22$ $161.22$ $c m$ $cm$

The z-score of

- 1.8

$-1.8$ is equivalent to a height of $161.22$ $161.22$ cm.

(a)

157.5

$157.5$ cm to

176.1

$176.1$ cm

(b) z-score:

2.16

$2.16$

161.22

$161.22$ cm

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1. Question

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$Z$ $Z$ score

Given Values

2. Question

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$Z$ $Z$ score

Given Values

3. Question

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$Z$ $Z$ score

4. Question

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$Z$ $Z$ score

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The Z Score

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ZZZ score

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ZZZ score

Given Values

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ZZZ score

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ZZZ score

Given Values

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$Z$ $Z$ score

$Z$ $Z$ score

$Z$ $Z$ score

$Z$ $Z$ score