Mathematics, 22.06.2021 01:00 leilaford2003

The college Physical Education Department offered an Advanced First Aid course last summer. The scores on the comprehensive final exam were normally distributed, and the z scores for some of the students are shown below. Robert, 1.11 Juan, 1.66 Susan, –1.9 Joel, 0.00 Jan, –0.65 Linda, 1.46
(a) Which of these students scored above the mean?
a. Jan
b. Joel
c. Juan
d. Linda
e. Robert
f. Susan
(b) Which of these students scored on the mean?
a. Jan
b. Joel
c. Juan
d. Linda
e. Robert
f. Susan
(c) Which of these students scored below the mean?
a. Jan
b. Joel
c. Juan
d. Linda
e. Robert
f. Susan
(d) If the mean score was ? = 156 with standard deviation ? = 24, what was the final exam score for each student? (Round your answers to the nearest whole number.)
a. Janb. Joelc. Juand. Lindae. Robertf. Susan

Answers: 3

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Answer from: only1123

b. Joel

c. Juan

d. Linda

b. Joel

a. Jan

f.Susan

a. Jan: 140

b. Joel: 156

c. Juan: 196

d. Linda: 191

e. Robert: 183

f. Susan: 110

Step-by-step explanation:

Z-score:

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean, positive z-scores are above the mean, negative are below the mean and 0 is the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Question a:

Robert, Juan and Linda had positive z-scores, so they scored above the mean, and the correct options are c,d,e.

(b) Which of these students scored on the mean?

Joel, which had a z-score of 0, so the correct option is b.

Jan and Susan had negative z-scores, so them, options a and f.

Question d:

We have that $\mu = 156, \sigma = 24$ , so we have to find X for each student.

Jan:

Z = -0.65. So

$Z = \frac{X - \mu}{\sigma}$

$-0.65 = \frac{X - 156}{24}$

X - 156 = -0.65*24

X = 140

b. Joel

Z = 0, so:

$Z = \frac{X - \mu}{\sigma}$

$0 = \frac{X - 156}{24}$

X - 156 = 0*24

X = 156

c. Juan

Z = 1.66, so:

$Z = \frac{X - \mu}{\sigma}$

$1.66 = \frac{X - 156}{24}$

X - 156 = 1.66*24

X = 196

d. Linda

Z = 1.46. So

$Z = \frac{X - \mu}{\sigma}$

$1.46 = \frac{X - 156}{24}$

X - 156 = 1.46*24

X = 191

e. Robert

Z = 1.11. So

$Z = \frac{X - \mu}{\sigma}$

$1.11 = \frac{X - 156}{24}$

X - 156 = 1.11*24

X = 183

f. Susan

Z = -1.9. So

$Z = \frac{X - \mu}{\sigma}$

$-1.9 = \frac{X - 156}{24}$

X - 156 = -1.9*24

X = 110

Answer from: Quest

439400

step-by-step explanation:

130^2 = 16900

16900*26 = 439400

Answer from: Quest

the length of o'p' is $1.5\times op$ .

step-by-step explanation:

it is given that the quadrilateral m'n'o'p' is the image of quadrilateral mnop after dilation.

the center of dilation is origin and the scale factor is 1.5.

in dilation the image and preimge are similar figures. it means the sides of image and preimage are proportional.

the scale factor is defined as

$k=\frac{o'p'}{op}$

the scale factor is 1.5.

$1.5=\frac{o'p'}{op}$

multiply both sides by op.

$1.5\times op=o'p'$

note: if the value of op is given, then multiply the value of op with 1.5 to get the length of side o'p'.

therefore length of o'p' is $1.5\times op$ .

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