Respuesta :
Answer:
(a) 0.2061
(b) 0.2514
(c) 0
Step-by-step explanation:
Let X denote the heights of women in the USA.
It is provided that X follows a normal distribution with a mean of 64 inches and a standard deviation of 3 inches.
(a)
Compute the probability that the sample mean is greater than 63 inches as follows:
[tex]P(\bar X>63)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{63-64}{3/\sqrt{6}})\\\\=P(Z>-0.82)\\\\=P(Z<0.82)\\\\=0.20611\\\\\approx 0.2061[/tex]
Thus, the probability that the sample mean is greater than 63 inches is 0.2061.
(b)
Compute the probability that a randomly selected woman is taller than 66 inches as follows:
[tex]P(X>66)=P(\frac{X-\mu}{\sigma}>\frac{66-64}{3})\\\\=P(Z>0.67)\\\\=1-P(Z<0.67)\\\\=1-0.74857\\\\=0.25143\\\\\approx 0.2514[/tex]
Thus, the probability that a randomly selected woman is taller than 66 inches is 0.2514.
(c)
Compute the probability that the mean height of a random sample of 100 women is greater than 66 inches as follows:
[tex]P(\bar X>66)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{66-64}{3/\sqrt{100}})\\\\=P(Z>6.67)\\\\\ =0[/tex]
Thus, the probability that the mean height of a random sample of 100 women is greater than 66 inches is 0.
