Given:
Mean, μ = 22
Standard deviation, σ = 7
Let's answer the following questions.
a. Given:
Sample size, n = 25
Let's find the probability that the sample mean is between 21.5 and 22.5.
We have:
![\begin{gathered} P(21.5Thus, we have:[tex]\begin{gathered} P(\frac{21.5-22}{\frac{7}{\sqrt[]{25}}}Using the standard normal table (NORMSDIST), we have:[tex]\begin{gathered} P(0.3571)=0.6395 \\ P(-0.3571)\text{ = }-0.3605 \\ \\ P(1.7857)-P(-0.3571)=0.6395-0.3605=0.279 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20P%2821.5Thus%2C%20we%20have%3A%5Btex%5D%5Cbegin%7Bgathered%7D%20P%28%5Cfrac%7B21.5-22%7D%7B%5Cfrac%7B7%7D%7B%5Csqrt%5B%5D%7B25%7D%7D%7DUsing%20the%20standard%20normal%20table%20%28NORMSDIST%29%2C%20we%20have%3A%5Btex%5D%5Cbegin%7Bgathered%7D%20P%280.3571%29%3D0.6395%20%5C%5C%20P%28-0.3571%29%5Ctext%7B%20%3D%20%7D-0.3605%20%5C%5C%20%20%5C%5C%20P%281.7857%29-P%28-0.3571%29%3D0.6395-0.3605%3D0.279%20%5Cend%7Bgathered%7D)
Therefore, the probability that sample mean is between 21.5 and 22.5 is 0.279
b. Given:
n = 25
Let's find the probability that the sample mean is between 21 and 22 minutes.
We have:
[tex]\begin{gathered} P(21Using the standard normal table, we have:[tex]\begin{gathered} P(-0.714286
Therefore, the probability that sample mean is between 21 and 22 is 0.2625
c. Given:
n = 144
Let's find the probability the sample mean is between 21.5 and 22.5
[tex]\begin{gathered} P(21.5
Therefore, the probability that sample mean is between 21.5 and 22.5 given a sample of 144 is 0.6086
d. Given:
Sample size in a = 25
Sample size in c = 144
The sample size in c is greater than the sample size in a so the standard error of the mean in (c) should be less than the standard error in (a).
As the standard error values become more concentrated
The average distance between the variable scores and the mean in a set of data is the standard deviation.
The answer is B.
Answer:
20
Step-by-step explanation:
1/10 of 200 is 20.
Answer:
cot(x)
Step-by-step explanation:



Recall the Pythagorean Identity 