Question

sample annual salaries (in thousands of dollars) for employees at a company are listed.
35 52 58 54 36 36 35 52 58 31 54 35 48
(a) find the sample mean and sample standard deviation.
(b) each employee in the sample is given a 5% raise. find the sample mean and sample standard deviation of the revised data set.
(c) to calculate the monthly salary, divide each original salary by 12. find the sample mean and sample standard deviation for the revised data set.
(d) what can you conclude from the results of (a), (b), and (c)?

(a) the sample mean is (\bar{x} = 44.9) thousand dollars.
(round to one decimal place as needed.)
the sample standard deviation is (s = square) thousand dollars.
(round to one decimal place as needed.)

Explanation:

Step1: Count the number of data points

First, we count the number of data points. Let's list the data: 35, 52, 58, 54, 36, 36, 35, 52, 58, 31, 54, 35, 48. Wait, there seems to be a typo in the original data (the "5" at the end? Maybe a missing number? Wait, let's count again. Wait, the original data: 35, 52, 58, 54, 36, 36, 35, 52, 58, 31, 54, 35, 48. Wait, that's 13 numbers? Wait, maybe the original data is 35, 52, 58, 54, 36, 36, 35, 52, 58, 31, 54, 35, 48 (13 data points). Wait, but let's confirm. Wait, maybe the user made a typo, but let's proceed with the given data. Wait, the sample mean is given as 44.9, so let's check the sum. Let's calculate the sum of the data:

35 + 52 + 58 + 54 + 36 + 36 + 35 + 52 + 58 + 31 + 54 + 35 + 48. Let's compute:

35*3 = 105; 52*2 = 104; 58*2 = 116; 54*2 = 108; 36*2 = 72; 31 + 48 = 79. Now sum these: 105 + 104 = 209; 209 + 116 = 325; 325 + 108 = 433; 433 + 72 = 505; 505 + 79 = 584. Wait, but the sample mean is 44.9, so 44.9 * n = sum. If n=13, 44.9*13 = 583.7, which is close (maybe a rounding error or a typo in the data). Let's assume n=13 (since there are 13 numbers listed, ignoring the last "5" which is probably a typo).

Step2: Calculate the deviations from the mean

Now, to find the sample standard deviation, we use the formula:

$$s = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{n - 1}}$$

First, we calculate each $(x_i - \bar{x})^2$:

For 35: $(35 - 44.9)^2 = (-9.9)^2 = 98.01$ (three times, so 3*98.01 = 294.03)

For 52: $(52 - 44.9)^2 = (7.1)^2 = 50.41$ (two times, so 2*50.41 = 100.82)

For 58: $(58 - 44.9)^2 = (13.1)^2 = 171.61$ (two times, so 2*171.61 = 343.22)

For 54: $(54 - 44.9)^2 = (9.1)^2 = 82.81$ (two times, so 2*82.81 = 165.62)

For 36: $(36 - 44.9)^2 = (-8.9)^2 = 79.21$ (two times, so 2*79.21 = 158.42)

For 31: $(31 - 44.9)^2 = (-13.9)^2 = 193.21$ (once)

For 48: $(48 - 44.9)^2 = (3.1)^2 = 9.61$ (once)

Now, sum all these squared deviations:

294.03 (35s) + 100.82 (52s) + 343.22 (58s) + 165.62 (54s) + 158.42 (36s) + 193.21 (31) + 9.61 (48)

Let's add step by step:

294.03 + 100.82 = 394.85

394.85 + 343.22 = 738.07

738.07 + 165.62 = 903.69

903.69 + 158.42 = 1062.11

1062.11 + 193.21 = 1255.32

1255.32 + 9.61 = 1264.93

Step3: Divide by (n - 1) and take the square root

Now, n = 13, so n - 1 = 12.

So, the variance is $\frac{1264.93}{12} \approx 105.4108$

Then, the standard deviation is $\sqrt{105.4108} \approx 10.3$ (rounded to one decimal place)

Wait, let's check the calculation again. Wait, maybe I made a mistake in the number of data points. Wait, let's recount the original data: 35, 52, 58, 54, 36, 36, 35, 52, 58, 31, 54, 35, 48. That's 13 numbers. Let's recalculate the sum:

35 + 35 + 35 = 105

52 + 52 = 104

58 + 58 = 116

54 + 54 = 108

36 + 36 = 72

31 + 48 = 79

Now sum: 105 + 104 = 209; 209 + 116 = 325; 325 + 108 = 433; 433 + 72 = 505; 505 + 79 = 584. Then the mean is 584 / 13 ≈ 44.923, which rounds to 44.9, so that's correct.

Now, recalculating the squared deviations:

For 35: (35 - 44.923)^2 ≈ (-9.923)^2 ≈ 98.466 (three times: 3*98.466 ≈ 295.398)

For 52: (52 - 44.923)^2 ≈ (7.077)^2 ≈ 50.084 (two times: 2*50.084 ≈ 100.168)

For 58: (58 - 44.923)^2 ≈ (13.077)^2 ≈ 171.008 (two times: 2*171.008 ≈ 342.016)

For 54: (54 - 44.923)^2 ≈ (9.077)^2 ≈ 82.392 (two times: 2*82.392 ≈ 164.784)

For 36: (36 - 44.923)^2 ≈ (-8.923)^2 ≈ 79.620 (two times: 2*79.620 ≈ 159.240)

For 31: (31 - 44.923)^2 ≈ (-13.923)^2 ≈ 193.850 (once)

For 48: (48 - 44.923)^2 ≈ (3.077)^2 ≈ 9.468 (once)

Now sum these:

295.398 + 100.168 = 395.566

395.566 + 342.016 = 737.582

737.582 + 164.784 = 902.366

902.366 + 159.240 = 1061.606

1061.606 + 193.850 = 1255.456

1255.456 + 9.468 = 1264.924

Now, divide by n - 1 = 12: 1264.924 / 12 ≈ 105.4103

Then, standard deviation is sqrt(105.4103) ≈ 10.3 (rounded to one decimal place)

Answer:

10.3