Q1 of company A = 2.5
Q3 of company A = 8
Interquatile range = (Q3 - Q1)/2 = (8 - 2.5)/2 = 5.5/2 = 2.75
Q1 of company B = 2
Q3 of company B = 5.5
Interquatile range = (5.5 - 2)/2 = 3.5/2 = 1.75
Therefore, t<span>he interquartile range for Company A employees is 2 more than the interquartile range for Company B employees.</span>
12+2x+1=59
13+2x=59
-13 - 13
2x =46
X = 23
You have to make a proportion:
12+x/8=8/x
Then cross multiply to solve for x
Answer:
The upper 20% of the weighs are weights of at least X, which is
, in which
is the standard deviation of all weights and
is the mean.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score of a measure X 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 from 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.
Upper 20% of weights:
The upper 20% of the weighs are weighs of at least X, which is found when Z has a p-value of 0.8. So X when Z = 0.84. Then
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![0.84 = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=0.84%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![X = 0.84\sigma + \mu](https://tex.z-dn.net/?f=X%20%3D%200.84%5Csigma%20%2B%20%5Cmu)
The upper 20% of the weighs are weights of at least X, which is
, in which
is the standard deviation of all weights and
is the mean.
Answer:
x=1
Step-by-step explanation:
hope this helps :)