Question

The state of California has a mean annual rainfall of 22 inches, whereas the state of New York has a mean annual rainfall of 42 inches ( Current Results website, October 27,2012). Assume that the standard deviation for both states is 4 inches. A sample of 30 years of rainfall for California and a sample of 45 years of rainfall for New York has been taken.

a. Show the probability distribution of the sample mean annual rainfall for California.

b. What is the probability that the sample mean is 1 inch of the population mean for California?

c. What is the probability that the sample mean is within 1 inch of the population mean for New York?

d. In which case, part (b) or (c), is the probability of obtaining a sample mean within 1 inch of the population mean greater? Why?

Answer 1

Answer:

a) Attached

b) P=0.84

c) P=0.89

d) Is greater in the case of the NY sample. This is because the bigger sample, which reduces the standard deviation of the sample mean.

Step-by-step explanation:

The probability distribution of the mean of the sample has this parameters:

$\mu_s=\mu_p\\\\\sigma_s=\sigma_p/\sqrt{n}$

The bigger the sample size, less spread of the distribution.

a) Show the probability distribution of the sample mean annual rainfall for California.

$\mu_s=\mu_p=22\\\\\sigma_s=\sigma_p/\sqrt{n}=4/\sqrt{30}=0.73$

The sampling distribution is attached.

b) What is the probability that the sample mean is 1 inch of the population mean for California?

We have to calculate the probability of having a sample mean between 21 and 23 inches.

We calculate the z values:

$z_1=\frac{x_1-\mu}{\sigma} =\frac{23-22}{0.73}=1.4\\\\z_2=\frac{x_2-\mu}{\sigma} =\frac{21-22}{0.73}=-1.4$

Both z-values are equidistant from the mean.

Now we can calculate the probability:

$P(|X-\mu|<1)=P(|z|<1.4)=0.84$

c) What is the probability that the sample mean is within 1 inch of the population mean for New York?

We have to calculate the sampling distribution parameters for the NY sample:

$\mu_s=\mu_p=42\\\\\sigma_s=\sigma_p/\sqrt{n}=4/\sqrt{45}=0.60$

The z-values are:

$z_1=\frac{x_1-\mu}{\sigma} =\frac{43-42}{0.60}=1.6\\\\z_2=\frac{x_2-\mu}{\sigma} =\frac{41-42}{0.60}=-1.6$

Now we can calculate the probability:

$P(|X-\mu|<1)=P(|z|<1.6)=0.89$

d) In which case, part (b) or (c), is the probability of obtaining a sample mean within 1 inch of the population mean greater? Why?

Is greater in the case of the NY sample. This is because the bigger sample, which reduces the standard deviation of the sample mean.

Bigger samples are expected, in average, to be more around the mean of the population.

Answer 2

Answer:

(A)The Probability distribution is attached in the solution.

(B) The Probability is 0.84 .

(C) The probability will be 0.89.

(D)The New York sample is getting higher probability within 1 inch mean population. This is because of the bigger sample size.

Step-by-step explanation:

Given information:

The probability distribution  $= \mu$

And, the standard deviation  $= \sigma$

(A) The probability distribution of mean annual rainfall in California.

$\mu_s=22\\\sigma_s=4/\sqrt30\\\sigma_s = 0.73$

The Probability distribution is attached in the solution.

(B) Now, we have to calculate the probability of having mean between 21 and 23 inches

Hence,

$Z_1=\frac{x_s-\mu}{\sigma} \\Z_1=(23-22)/0.73\\Z_1=1.4$

Similarly ,

$Z_2=(21-22)/0.73\\Z_2=-1.4$

Now, calculate the probability as;

$P(|X-\mu |<1)=P(|z|<1.4) = 0.84$

(C)The sample mean is within 1 inch of the population mean of New York:

For this ;

$\mu_s=42\\\sigma_s=\sigma_p/n\\\sigma_s=4/\sqrt45\\\sigma_s=0.60$

Now, calculate Z values:

$Z_1=(43-42)/0.80\\Z_1=1.6$

Similarly:

$Z_2=(41-42)/0.60\\Z_2=-1.6$

Hence, the probability will be:

$P(|X-\mu |<1)=P(|z|<1.6) = 0.89$

(D)The New York sample is getting higher probability within 1 inch mean population. This is because of the bigger sample size, Which reduces the value of standard deviation.

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