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Answer:
The height that is at the 20th percentile is 62.9 in.
Step-by-step explanation:
A percentile indicates the value which a given percentage of observations in a group of observations falls.
The 20th percentile is the value below which 20% of the observations may be found.
In this case, we have a normal distribution with mean of 65 in. and standard deviation of 2.5 in.
We can use the standarized z-table to find the value of z for the 20th percentile. The value of z for the 20th percentile is z=-0.842.
We can transform this to the normal distribution of this model ([tex]N(65,2.5^2)[/tex]):
[tex]x=\mu+z*\sigma=65+(-0.842)*2.5=62.9[/tex]
The height that is at the 20th percentile is 62.9 in.
In the picture we can see the percentage that falls below this value (20%).
The height at the 20th percentile of the data set considered for heights of women at the considered college is 62.9 inches approx
How to get the z scores?
If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z-score.
If we have
[tex]X \sim N(\mu, \sigma)[/tex]
(X is following normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] )
then it can be converted to standard normal distribution as
[tex]Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)[/tex]
(Know the fact that in continuous distribution, probability of a single point is 0, so we can write
[tex]P(Z \leq z) = P(Z < z) )[/tex]
Also, know that if we look for Z = z in z tables, the p value we get is
[tex]P(Z \leq z) = \rm p \: value[/tex]
How to convert percent to probability?
Percent counts the number compared to 100 whereas probability counts it compare to 1.
So, if we have a%, that means for each 100, there are 'a' parts. If we divide each of them with 100, we get:
For each 1, there are a/100 parts.
Thus, 50% = 50/100 = 0.50 (in probability)
For the considered case, let the random variable X assumes the data of heights of the women of the considered college. Then, by the given information, we get:
[tex]X \sim N(\mu = 65, \sigma = 2.5)[/tex] (data is in inches)
Now, let the height at 20th percentile be X = x inches.
Then, that means
[tex]P(X \leq x) = 20\% = 0.2[/tex]
Converting the current distribution of X to standard normal distribution, we get:
[tex]Z = \dfrac{X - \mu}{\sigma}\\[/tex]
or
[tex]P(X \leq x) = 20\% = 0.2\\P(Z \leq z = \dfrac{x- \mu}{\sigma}) = P(Z \leq z = \dfrac{x - 65}{2.5}) = 0.2\\[/tex]
From the z-tables, the value of z for which the p-value is obtained approx 0.2 is Z = -0.84
Thus, we get:
[tex]Z = z\approx-0.84\\\\\dfrac{x-65}{2.5} \approx -0.84\\\\x \approx 65 - 2.1 =62.9 \: \rm inches[/tex]
Thus, the height at the 20th percentile of the data set considered for heights of women at the considered college is 62.9 inches approx
Learn more about z-score here:
https://brainly.com/question/21262765
