Question

1. a random sample means that

a only volunteers are selected
b every member of the population has an equal chance of being selected
c the first few members of a group are selected
d the largest or smallest values are selected

1. mia recorded the number of pages she read each day for a week: 25, 30, 15, 20, 35, 40, 25. what is the mean absolute deviation of this data?

Explanation:

Step1: Find the mean of the data set

The data set is \(25, 30, 15, 20, 35, 40, 25\). First, calculate the sum of the data points: \(25 + 30 + 15 + 20 + 35 + 40 + 25 = 190\). There are 7 data points, so the mean \(\bar{x}=\frac{190}{7}\approx27.14\) (we can also keep it as a fraction \(\frac{190}{7}\) for accuracy). Wait, wait, let's recalculate the sum: \(25+30 = 55\), \(55 + 15 = 70\), \(70 + 20 = 90\), \(90 + 35 = 125\), \(125 + 40 = 165\), \(165 + 25 = 190\). Yes, sum is 190, number of data points \(n = 7\), so mean \(\bar{x}=\frac{190}{7}\approx27.14\) (or exactly \(\frac{190}{7}\)).

Step2: Calculate the absolute deviation for each data point

For each \(x_i\), find \(|x_i - \bar{x}|\):
- For \(x = 25\): \(|25 - \frac{190}{7}|=|\frac{175 - 190}{7}|=|\frac{-15}{7}|=\frac{15}{7}\approx2.14\)
- For \(x = 30\): \(|30 - \frac{190}{7}|=|\frac{210 - 190}{7}|=|\frac{20}{7}|\approx2.86\)
- For \(x = 15\): \(|15 - \frac{190}{7}|=|\frac{105 - 190}{7}|=|\frac{-85}{7}|=\frac{85}{7}\approx12.14\)
- For \(x = 20\): \(|20 - \frac{190}{7}|=|\frac{140 - 190}{7}|=|\frac{-50}{7}|=\frac{50}{7}\approx7.14\)
- For \(x = 35\): \(|35 - \frac{190}{7}|=|\frac{245 - 190}{7}|=|\frac{55}{7}|\approx7.86\)
- For \(x = 40\): \(|40 - \frac{190}{7}|=|\frac{280 - 190}{7}|=|\frac{90}{7}|\approx12.86\)
- For \(x = 25\): \(|25 - \frac{190}{7}|=\frac{15}{7}\approx2.14\) (same as the first 25)

Step3: Find the mean of these absolute deviations

Sum of absolute deviations: \(\frac{15}{7}+\frac{20}{7}+\frac{85}{7}+\frac{50}{7}+\frac{55}{7}+\frac{90}{7}+\frac{15}{7}=\frac{15 + 20 + 85 + 50 + 55 + 90 + 15}{7}=\frac{330}{7}\approx47.14\) (wait, no, wait: 15+20=35, 35+85=120, 120+50=170, 170+55=225, 225+90=315, 315+15=330. Yes, sum is 330. Then mean absolute deviation (MAD) is \(\frac{330}{7}\approx47.14\)? Wait, that can't be right. Wait, maybe I made a mistake in the data set. Wait, the problem says "25, 30, 15, 20, 35, 40, 25" – wait, let's check the original data again. The user wrote: "25, 30, 15, 20, 35, 40, 25" (wait, the original text: "25, 30, 15, 20, 35, 40, 25" – wait, maybe I miscounted the data points. Wait, the problem says "for a week" – 7 days. Let's list the data again: 25, 30, 15, 20, 35, 40, 25. Let's recalculate the sum: 25+30=55, +15=70, +20=90, +35=125, +40=165, +25=190. Correct. Mean is 190/7 ≈27.14. Now, absolute deviations:

25: |25 - 27.14| = 2.14

30: |30 - 27.14| = 2.86

15: |15 - 27.14| = 12.14

20: |20 - 27.14| = 7.14

35: |35 - 27.14| = 7.86

40: |40 - 27.14| = 12.86

25: |25 - 27.14| = 2.14

Now sum these deviations: 2.14 + 2.86 = 5; 5 + 12.14 = 17.14; 17.14 + 7.14 = 24.28; 24.28 + 7.86 = 32.14; 32.14 + 12.86 = 45; 45 + 2.14 = 47.14. Then divide by 7: 47.14 / 7 ≈6.73? Wait, no, wait 330/7 is approximately 47.14, and 47.14 divided by 7 is approximately 6.73? Wait, no, 330 divided by 7 is 47.14, and 47.14 divided by 7 is 6.73? Wait, no, 330/7 is the sum of absolute deviations, then MAD is (330/7)/7? No, no! Wait, mean absolute deviation is the mean of the absolute deviations, so sum of absolute deviations divided by the number of data points (n=7). So sum of absolute deviations is 330/7, then divide by 7? No, no! Wait, no: each absolute deviation is |x_i - x̄|, then sum all |x_i - x̄|, then divide by n. So sum of |x_i - x̄| is (15 + 20 + 85 + 50 + 55 + 90 + 15)/7 = 330/7. Then MAD is (330/7) / 7? No, no! Wait, n is 7, so sum of absolute deviations is 330/7? No, wait, 15/7 + 20/7 + 85/7 + 50/7 + 55/7 + 90/7 + 15/7 = (15+20+85+50+55+90+15)/7 = 330/7. Then MAD is (330/7) / 7? No, that's wrong. Wait, no: the formula for MAD is \(\frac{1}{n}\sum_{i = 1}^{n}|x_i - \bar{x}|\). So n=7, sum of |x_i - x̄| is 330/7? No, wait, 15/7 is approximately 2.14, 20/7≈2.86, 85/7≈12.14, 50/7≈7.14, 55/7≈7.86, 90/7≈12.86, 15/7≈2.14. Adding these decimals: 2.14 + 2.86 = 5; 5 + 12.14 = 17.14; 17.14 + 7.14 = 24.28; 24.28 + 7.86 = 32.14; 32.14 + 12.86 = 45; 45 + 2.14 = 47.14. Then 47.14 divided by 7 is approximately 6.73. Wait, but let's do it with fractions. Sum of absolute deviations: 15 + 20 + 85 + 50 + 55 + 90 + 15 = 330. So sum of absolute deviations is 330, then divide by n=7. Wait, oh! I see my mistake. Earlier, I converted each absolute deviation to a fraction with denominator 7, but actually, |x_i - x̄| can be calculated as |x_i - (190/7)|, but when we sum them, we can also calculate the absolute differences first without converting to sevenths. Let's recalculate the absolute differences correctly:

x̄ = 190/7 ≈27.1429

|25 - 27.1429| = 2.1429

|30 - 27.1429| = 2.8571

|15 - 27.1429| = 12.1429

|20 - 27.1429| = 7.1429

|35 - 27.1429| = 7.8571

|40 - 27.1429| = 12.8571

|25 - 27.1429| = 2.1429

Now sum these: 2.1429 + 2.8571 = 5; 5 + 12.1429 = 17.1429; 17.1429 + 7.1429 = 24.2858; 24.2858 + 7.8571 = 32.1429; 32.1429 + 12.8571 = 45; 45 + 2.1429 = 47.1429. Now divide by 7: 47.1429 / 7 ≈6.7347. So approximately 6.73, or as a fraction, 330/49 ≈6.73.

Wait, but maybe I made a mistake in the data set. Let's check the original data again: the user wrote "25, 30, 15, 20, 35, 40, 25" – yes, 7 numbers. Let's recalculate the mean: (25 + 30 + 15 + 20 + 35 + 40 + 25) = 25*2 + 30 + 15 + 20 + 35 + 40 = 50 + 30 = 80 + 15 = 95 + 20 = 115 + 35 = 150 + 40 = 190. Mean is 190/7 ≈27.14. Then absolute deviations:

25: 27.14 - 25 = 2.14

30: 30 - 27.14 = 2.86

15: 27.14 - 15 = 12.14

20: 27.14 - 20 = 7.14

35: 35 - 27.14 = 7.86

40: 40 - 27.14 = 12.86

25: 27.14 - 25 = 2.14

Sum: 2.14 + 2.86 = 5; +12.14 = 17.14; +7.14 = 24.28; +7.86 = 32.14; +12.86 = 45; +2.14 = 47.14. Divide by 7: 47.14 / 7 ≈6.73. So the mean absolute deviation is approximately 6.73.

Answer:

The mean absolute deviation is approximately \(\boxed{6.73}\) (or as a fraction \(\frac{330}{49}\) or approximately 6.7).