Which is not a representation of the equation y = 0.5x

Heights of Women. Heights of adult women are distributed normally with a mean of 162 centimeters and a standard deviation of 8 c

Bogdan [553]

Answer:

a) 6.68% of heights less than 150 centimeters

b) 58.65% of heights between 160 centimeters and 180 centimeters

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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. 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 pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

$\mu = 162, \sigma = 8$

a) The percentage of heights less than 150 centimeters

We have to find the pvalue of Z when X = 150. So

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

$Z = \frac{150 - 162}{8}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.0668

6.68% of heights less than 150 centimeters

b) The percentage of heights between 160 centimeters and 180 centimeters

We have to find the pvalue of Z when X = 180 subtracted by the pvalue of Z when X = 160.

X = 180

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

$Z = \frac{180 - 162}{8}$

$Z = 2.25$

$Z = 2.25$ has a pvalue of 0.9878

X = 160

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

$Z = \frac{160 - 162}{8}$

$Z = -0.25$

$Z = -0.25$ has a pvalue of 0.4013

0.9878 - 0.4013 = 0.5865

58.65% of heights between 160 centimeters and 180 centimeters

7 0

3 years ago

NEED ALL 4! With the solution and step by step on what to do !! worth 35 points

creativ13 [48]

Answer:

I got you, give me a few minutes

1st one: 6x+12+6y

2nd one: -12a-20b

4th one: 2x times 18

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

an outlier may be defined as a data point that is more than 1.5 times the interquartile range below the lower quartile or is mor

Ira Lisetskai [31]

Supposing a normal distribution, we find that:

The diameter of the smallest tree that is an outlier is of 16.36 inches.

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We suppose that tree diameters are normally distributed with mean 8.8 inches and standard deviation 2.8 inches.

In a normal distribution 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.
After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem: