Respuesta :
Answer:
a) Sampling distribution of sample means; b) The mean, μ, is 20.8 and the standard error, [tex]\sigma_{\bar{x}}[/tex], is 0.3692; c) 0.0034; d) 0.9994; e) 20.55626.
Step-by-step explanation:
For part a,
The name of a distribution of sample means is the sampling distribution of sample means. The mean of a distribution of sample means is the same as the population mean, μ, which is 20.8. The standard error of a sampling distribution of sample means is represented by [tex]\sigma_{\bar{x}}[/tex], and is given by σ/√n:
4.8/√169 = 4.8/13 = 0.3692.
For part c,
We use the formula for a z score,
[tex]z=\frac{\bar{X}-\mu}{\sigma \div \sqrt{n}}[/tex]
z = (21.8-20.8)/0.3692 = 1/0.3692 = 2.71
Using a z table, we see that the area under the curve to the left of this is 0.9966. However, we want the area under the curve to the right, so we subtract from 1:
1-0.9966 = 0.0034
For part d,
We want P(19.3 ≤ X ≤ 22):
z = (19.3-20.8)/0.3692 = -1.5/0.3692 = -4.06
z = (22-20.8)/0.3692 = 1.2/0.3692 = 3.25
The area under the curve to the right of z = -4.06 is 0.00. The area under the curve to the right of z = 3.25 is 0.9994. This makes the area between them
0.9994 - 0.00 = 0.9994.
For part e,
We look up 25%, or 0.25, in the z table. The closest value to this is 0.2514, which corresponds to a z score of -0.67:
-0.67 = (X-20.8)/0.3692
Multiply both sides by 0.3692:
0.3692(-0.67) = ((X-20.8)/0.3692)(0.3692)
-0.247364 = X-20.8
Add 20.8 to each side:
-0.247364+20.8 = X-20.8+20.8
20.5526 = X
