Respuesta :
(Hope that helps (= The given says the height distribution is normally distributed with the mean height equal to 68 inches. In this case, the bell-shaped curve has a vertical symmetry at 68 inches. This means, half of the mean exceeds 68 inches while the other half has height below 68 inches.
The height range out of which only 5% of the young American men (from age 18 to 24) lie is [63.35, 72.65] (in inches)
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]
For this case, let we take:
- X = height of American men from age 18 to 24
- [a,b] = range of height (the values of X) outside which there lies only 5% of American men.
Then, according to the given data, we have:
[tex]X \sim N(\mu = 68, \sigma = 2.5)[/tex]
where [tex]\mu[/tex] is mean value of X and [tex]\sigma[/tex] is standard deviation of X (both in inches).
Also, we can write:
[tex]P(X < a) + P(X > b) = 5\% = 0.05[/tex]
Since normal distribution is symmetric about its mean, we can take a and b equidistant from the mean, so as to get a symmetric range which makes much more sense than taking an asymmetric range which doesn't comply with the nature of values of X.
Thus, we have:
[tex]\mu - a = b - \mu\\b = 2\mu - a[/tex]
From this result and [tex]P(X < a) + P(X > b) = 5\% = 0.05[/tex], we get:
[tex]P(X < a) + P(X > 2\mu -a) = 0.05[/tex]
Converting X to Z(the standard normal distribution), we get;
[tex]Z = \dfrac{X - \mu}{\sigma}[/tex]
[tex]P(X < a) + P(X > 2\mu -a) = 0.05\\\\P(Z < z = \dfrac{x-\mu}{\sigma} = \dfrac{a-68}{2.5}) + P(Z > \dfrac{2(68) - a - 68}{2.5}) = 0.05\\\\P(Z < \dfrac{a-68}{2.5}) + P(Z > \dfrac{68-a}{2.5}) = 0.05\\\\P(Z < \dfrac{a-68}{2.5}) + P(Z > -\dfrac{a-68}{2.5}) = 0.05\\\\P(Z < \dfrac{a-68}{2.5}) + P(Z \leq \dfrac{a-68}{2.5}) = 0.05 \: \: \: \: (\because P(Z > -k) = P(Z \leq k))\\\\2P(Z < \dfrac{a-68}{2.5}) = 0.05\\\\P(Z < \dfrac{a-68}{2.5}) = 0.025[/tex]
Using the z-tables, we get the value of Z for which p-value is 0.025 as
-1.96
Thus, we get:
[tex]\dfrac{a-68}{2.5} = -1.96\\\\a = 68 + (-1.96 \times 2.5)\\a = 63.35[/tex]
Thus, we get: [tex]b = 2\mu - a = 2(68) - 63.35 = 72.65[/tex]
Thus, the range is [a,b] = [63.35, 72.65]
Thus, the height range out of which only 5% of the young American men (from age 18 to 24) lie is [63.35, 72.65] (in inches)
Learn more about standard normal distribution here:
https://brainly.com/question/10984889
