Answer:
13. 200
14. 8,300
15. 500
16. 2,500
Step-by-step explanation:
13. It is more than 4
14. It is more than 4
15. It is less than 4
16. It is more than 4
Answer:
Step-by-step explanation:
Since 40% of the crew are men, which means n. 40=8 n=8/. 40 n=20 total number of crew 20 - 8=12 are women
Answer:
x = 2.5
Step-by-step explanation:
2x - 11 = -8x + 14
Add 8x to both sides
10x - 11 = 14
Add 11 to both sides
10x = 25
Divide both sides by 10
x = 2.5
1: 0.5 and 50%
2: .2 and 20%
3: .75 and 75%
10: (1/10) and 10%
11: (3/5) and 60%
12: (1/4) and 25%
Using the normal distribution, it is found that there are 68 students with scores between 72 and 82.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean
and standard deviation
is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
In this problem, the mean and the standard deviation are given, respectively, by:
![\mu = 72, \sigma = 10](https://tex.z-dn.net/?f=%5Cmu%20%3D%2072%2C%20%5Csigma%20%3D%2010)
The proportion of students with scores between 72 and 82 is the <u>p-value of Z when X = 82 subtracted by the p-value of Z when X = 72</u>.
X = 82:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{82 - 72}{10}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B82%20-%2072%7D%7B10%7D)
Z = 1
Z = 1 has a p-value of 0.84.
X = 72:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{72 - 72}{10}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B72%20-%2072%7D%7B10%7D)
Z = 0
Z = 0 has a p-value of 0.5.
0.84 - 0.5 = 0.34.
Out of 200 students, the number is given by:
0.34 x 200 = 68 students with scores between 72 and 82.
More can be learned about the normal distribution at brainly.com/question/24663213
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