# I need help

28) The ages (in years) of the four U.S. vice presidents who assumed office after presidential assassinations are 56 (A. Johnson), 51 (C. Arthur), 42 (T. Roosevelt), and 55 (L.B. Johnson). (i) Assuming that 2 of the ages are randomly selected with replacement, list the 16 different possible samples. (ii) Find the standard deviation of each of the 16 samples, then summarize the sampling distribution of the standard deviation in the format of a table representing the probability distribution. (iii) Compare the population standard deviation to the mean of the sample standard deviations. (iv) Do the sample standard deviations target the value of the population standard deviation? In general, do sample standard deviations make good estimators of population standard deviations? Why or why not?

27) The weekly salaries of the teachers in one state are normally distributed with a mean of $490 and a standard deviation of$45. What is the probability that a randonmly selected teacher earns more than $525 a week (Use TI-84) for this problem. asked Mar 17, 2015 ## 3 Answers 0 votes Step 1: (27) Let the random variable x be the weekly salaries of the teacher in a state. The mean of random variable =$ 490.

The standard deviation of random variable = $45. Find the probability of a teacher earning more than$ 525 a week.

P(x > 525).

Now calculate the Z-score for the given probability.

,

where x is the weekly salaries of the teacher,

is the mean and

is the standard deviation.

For x = 525, Z-score is Z = (525 - 490)/45

From the Z-score table, Area of the region when Z = 0.78 is 0.7823.

P(Z > 0.78)

= 1 - P(Z < 0.78)

= 1-[Area of the region when Z = 0.78]

= 1 - [0.7823]

= 0.2177

The probability of weekly salaries of the teacher greater than $525 is 0.2177. Solution: (27) The probability of weekly salaries of the teacher greater than$525 is 0.2177.

Using TI-84 Calculator,

You need to perform the following operations in your calculator as shown below.

normalcdf(0.7823, 100)

Then you will get P(x > 525) = P(Z > 0.78) = 0.2177.

edited Mar 17, 2015 by Lucy
+1 vote

Step 1:

(28)

(i)

The ages of the four U.S. vice presidents are 56 , 51, 42 and 55.

Two members are selected randomly.

The 16 different possible samples are

{ (56, 56), (56, 51), (56, 42), (56, 55), (51, 56), (51, 51), (51, 42), (51, 55),

(42, 56), (42, 51), (42, 42), (42, 55), (55, 56), (55, 51), (55, 42), (55, 55)}

Step 2:

(ii)

Complete the table.

Write down the samples and find the standard deviation.

Mean = $\fn_cm \small \left ( \frac {1}{n}\sum x \right )$.

= (1/16) (816)

= 51

Mean µ = 51.

Observe the table:

Variance σ² =  ∑(xµ)²P(x)=15.26

Standard deviation σ = √ 15.26 = 3.90

Standard deviation for each sample is 3.90.

Solution:

(i)

The 16 different possible samples are

{ (56, 56), (56, 51), (56, 42), (56, 55), (51, 56), (51, 51), (51, 42), (51, 55),

(42, 56), (42, 51), (42, 42), (42, 55), (55, 56), (55, 51), (55, 42), (55, 55)}

(ii)

Standard deviation for each sample is 3.90.

edited Mar 17, 2015 by Lucy

Step 1:

(28)

(iii)

Population Standard deviation σ = 3.90        (From (28)(ii))

Sample standard deviation

$\fn_cm \small \fn_cm \small \sigma = \sqrt{\frac{(x-\mu )^2}{n-1}}$

σ = √(244/15) = 4.033

Sample standard deviation σ = 4.033.

Population standard deviation is 3.90 and sample standard deviation is 4.03.

They are not equal.

Step 2:

(iv)

The sample standard deviation do not target the population standard deviation.

So sample standard deviation do not make good estimators of population standard deviation.

Solution:

(iii)

Population standard deviation is 3.90 and sample standard deviation is 4.03.

They are not equal.

(iv)

The sample standard deviation do not target the population standard deviation.