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Question 1 of 4
If a set of scores have a mean of 11 and a standard deviation of 0.9, sketch the normal curve then find the z-score for 12.
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Given Values
Mean ¯¯¯X=11
Standard Deviation=0.9
First, complete the labels of the bell curve by using the mean and standard deviation.
For example, start with the mean,
11. Then add and subtract
0.9 to get the values
1 standard deviation above and below the mean.
Keep adding and subtracting the standard deviation until the labels are completed.
Input the values to the z-score formula to solve for the z-score.
Z |
= |
x−¯xS |
|
|
= |
12−110.9 |
Substitute values |
|
|
= |
1.11 |
Simplify |
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Question 2 of 4
Given that the mean is 78.4 and the standard deviation is 3.2, find the z-scores for 68.2 and 82.4.
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Given Values
Mean ¯¯¯X=78.4
Standard Deviation=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. Then add and subtract
3.2 to get the values
1 standard deviation above and below the mean.
Keep adding and subtracting the standard deviation until the labels are completed.
Input the values to the z-score formula to solve for the z-score of 68.2.
Z |
= |
x−¯xS |
|
|
= |
68.2−78.43.2 |
Substitute values |
|
|
= |
−10.23.2 |
Simplify |
|
|
= |
−3.19 |
Do the same to solve for the z-score of 82.4.
Z |
= |
x−¯xS |
|
|
= |
82.4−78.43.2 |
Substitute values |
|
|
= |
43.2 |
Simplify |
|
|
= |
1.25 |
68.2 z-score: −3.19
82.4 z-score: 1.25
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Question 3 of 4
Jack had a test for both Math and English. Given the following information, in which of the two subjects did he perform better?
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First, find the z-score for Math.
Z |
= |
x−¯xS |
|
|
= |
86−6213 |
Substitute values |
|
|
= |
2413 |
|
|
= |
1.85 |
First, find the z-score for English.
Z |
= |
x−¯xS |
|
|
= |
86−736 |
Substitute values |
|
|
= |
136 |
|
|
= |
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
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Question 4 of 4
Women’s heights in a survey had a mean of 166.8cm, a standard deviation of 3.1cm, and is normally distributed.
(a) Where do most heights almost certainly lie?
(b) Find the z-score for the height of 173.5cm.
(c) Find the height for the z-score of −1.8.
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Given Values
Mean ¯¯¯X=166.8 cm
Standard Deviation=3.1 cm
First, complete the labels of the bell curve by using the mean and standard deviation.
For example, start with the mean, 166.8 cm. Then add and subtract 3.1 cm to get the values 1 standard deviation above and below the mean.
Keep adding and subtracting the standard deviation until the labels are completed.
99.7% of scores lie within 3 standard deviations away from the mean.
Hence, almost all heights measure 157.5 cm to 176.1 cm
For part (b), input the values to the z-score formula to solve for the z-score of the height, 173.5 cm.
Z |
= |
x−¯xS |
|
|
= |
173.5−166.83.1 |
Substitute values |
|
|
= |
6.73.1 |
Simplify |
|
|
= |
2.16 |
The actual height of 173.5cm is equivalent to a z-score of 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.
Z |
= |
x−¯xS |
|
−1.8 |
= |
x−166.83.1 |
Substitute values |
|
−1.8×3.1 |
= |
x−166.83.1×3.1 |
Multiply both sides by 3.1 |
|
−5.58 |
= |
x−166.8 |
Simplify |
|
−5.58+166.8 |
= |
x−166.8+166.8 |
Add 166.8 to both sides |
161.22 |
= |
x |
x |
= |
161.22 cm |
The z-score of −1.8 is equivalent to a height of 161.22cm.
(a) 157.5cm to 176.1cm
(b) z-score: 2.16
(c) height: 161.22cm