Question

58 randomly selected students were asked the number of pairs of shoes they have. let x represent number of pairs of shoes. the results are as follows:

of pairs of shoes | frequency

4 | 5
5 | 4
6 | 6
7 | 6
8 | 3
9 | 4
10 | 7
11 | 5
12 | 6
13 | 3
14 | 6
15 | 3

round all your answers to at least 2 decimal places where possible.

the mean is: 9.36
the median is: 10.00
the sample standard deviation is: 4.29
the first quartile is: 6.00
the third quartile is: 12.00

Explanation:

Step1: Calculate the midpoint and product for each class

First, we need to find the midpoint \( x \) for each number of pairs of shoes (since it's a discrete data, \( x \) is the number of pairs itself). Then calculate \( f \cdot x \) (frequency times \( x \)) and \( f \cdot x^2 \) (frequency times \( x \) squared) for each row.

| \( x \) | \( f \) | \( f \cdot x \) | \( f \cdot x^2 \) |
|--------|--------|----------------|------------------|
| 4 | 5 | \( 5 \times 4 = 20 \) | \( 5 \times 16 = 80 \) |
| 5 | 4 | \( 4 \times 5 = 20 \) | \( 4 \times 25 = 100 \) |
| 6 | 6 | \( 6 \times 6 = 36 \) | \( 6 \times 36 = 216 \) |
| 7 | 6 | \( 6 \times 7 = 42 \) | \( 6 \times 49 = 294 \) |
| 8 | 3 | \( 3 \times 8 = 24 \) | \( 3 \times 64 = 192 \) |
| 9 | 4 | \( 4 \times 9 = 36 \) | \( 4 \times 81 = 324 \) |
| 10 | 7 | \( 7 \times 10 = 70 \) | \( 7 \times 100 = 700 \) |
| 11 | 5 | \( 5 \times 11 = 55 \) | \( 5 \times 121 = 605 \) |
| 12 | 6 | \( 6 \times 12 = 72 \) | \( 6 \times 144 = 864 \) |
| 13 | 3 | \( 3 \times 13 = 39 \) | \( 3 \times 169 = 507 \) |
| 14 | 6 | \( 6 \times 14 = 84 \) | \( 6 \times 196 = 1176 \) |
| 15 | 3 | \( 3 \times 15 = 45 \) | \( 3 \times 225 = 675 \) |

Step2: Calculate the sum of \( f \), \( f \cdot x \), and \( f \cdot x^2 \)

Sum of frequencies \( \sum f = 5 + 4 + 6 + 6 + 3 + 4 + 7 + 5 + 6 + 3 + 6 + 3 = 58 \) (which matches the given number of students).

Sum of \( f \cdot x \): \( \sum (f \cdot x) = 20 + 20 + 36 + 42 + 24 + 36 + 70 + 55 + 72 + 39 + 84 + 45 = 543 \)

Sum of \( f \cdot x^2 \): \( \sum (f \cdot x^2) = 80 + 100 + 216 + 294 + 192 + 324 + 700 + 605 + 864 + 507 + 1176 + 675 = 5737 \)

Step3: Calculate the sample variance

The formula for sample variance \( s^2 \) is:
\[ s^2 = \frac{\sum (f \cdot x^2) - \frac{(\sum (f \cdot x))^2}{n}}{n - 1} \]
where \( n = \sum f = 58 \)

First, calculate \( \frac{(\sum (f \cdot x))^2}{n} = \frac{543^2}{58} = \frac{294849}{58} \approx 5083.6034 \)

Then, \( \sum (f \cdot x^2) - \frac{(\sum (f \cdot x))^2}{n} = 5737 - 5083.6034 = 653.3966 \)

Now, divide by \( n - 1 = 57 \):
\[ s^2 = \frac{653.3966}{57} \approx 11.4631 \]

Step4: Calculate the sample standard deviation

The sample standard deviation \( s \) is the square root of the sample variance:
\[ s = \sqrt{11.4631} \approx 3.3857 \approx 3.39 \]

Wait, let's check the calculations again. Wait, maybe I made a mistake in the sum of \( f \cdot x \) or \( f \cdot x^2 \). Let's recalculate the sum of \( f \cdot x \):

4*5=20; 5*4=20 (total 40); 6*6=36 (76); 7*6=42 (118); 8*3=24 (142); 9*4=36 (178); 10*7=70 (248); 11*5=55 (303); 12*6=72 (375); 13*3=39 (414); 14*6=84 (498); 15*3=45 (543). That's correct.

Sum of \( f \cdot x^2 \):

4²*5=16*5=80; 5²*4=25*4=100 (180); 6²*6=36*6=216 (396); 7²*6=49*6=294 (690); 8²*3=64*3=192 (882); 9²*4=81*4=324 (1206); 10²*7=100*7=700 (1906); 11²*5=121*5=605 (2511); 12²*6=144*6=864 (3375); 13²*3=169*3=507 (3882); 14²*6=196*6=1176 (5058); 15²*3=225*3=675 (5733). Wait, earlier I had 5737, that was a mistake. Let's recalculate:

80 + 100 = 180; 180 + 216 = 396; 396 + 294 = 690; 690 + 192 = 882; 882 + 324 = 1206; 1206 + 700 = 1906; 1906 + 605 = 2511; 2511 + 864 = 3375; 3375 + 507 = 3882; 3882 + 1176 = 5058; 5058 + 675 = 5733. Ah, there was a mistake in the previous sum of \( f \cdot x^2 \). It should be 5733, not 5737.

So now, \( \frac{(\sum (f \cdot x))^2}{n} = \frac{543^2}{58} = \frac{294849}{58} \approx 5083.6034 \)

Then, \( \sum (f \cdot x^2) - \frac{(\sum (f \cdot x))^2}{n} = 5733 - 5083.6034 = 649.3966 \)

Now, divide by \( n - 1 = 57 \):

\( s^2 = \frac{649.3966}{57} \approx 11.3929 \)

Then, sample standard deviation \( s = \sqrt{11.3929} \approx 3.375 \approx 3.38 \)

Wait, maybe the initial approach is wrong? Wait, the data is discrete, so the formula for sample standard deviation when data is in frequency distribution is:

\( s = \sqrt{\frac{\sum f(x - \bar{x})^2}{n - 1}} \)

We know that \( \bar{x} = \frac{\sum fx}{n} = \frac{543}{58} \approx 9.3621 \)

Let's calculate \( \sum f(x - \bar{x})^2 \):

For each \( x \):

- \( x=4 \): \( 5 \times (4 - 9.3621)^2 = 5 \times (-5.3621)^2 = 5 \times 28.752 = 143.76 \)
- \( x=5 \): \( 4 \times (5 - 9.3621)^2 = 4 \times (-4.3621)^2 = 4 \times 19.028 = 76.112 \)
- \( x=6 \): \( 6 \times (6 - 9.3621)^2 = 6 \times (-3.3621)^2 = 6 \times 11.303 = 67.818 \)
- \( x=7 \): \( 6 \times (7 - 9.3621)^2 = 6 \times (-2.3621)^2 = 6 \times 5.580 = 33.48 \)
- \( x=8 \): \( 3 \times (8 - 9.3621)^2 = 3 \times (-1.3621)^2 = 3 \times 1.855 = 5.565 \)
- \( x=9 \): \( 4 \times (9 - 9.3621)^2 = 4 \times (-0.3621)^2 = 4 \times 0.131 = 0.524 \)
- \( x=10 \): \( 7 \times (10 - 9.3621)^2 = 7 \times (0.6379)^2 = 7 \times 0.407 = 2.849 \)
- \( x=11 \): \( 5 \times (11 - 9.3621)^2 = 5 \times (1.6379)^2 = 5 \times 2.683 = 13.415 \)
- \( x=12 \): \( 6 \times (12 - 9.3621)^2 = 6 \times (2.6379)^2 = 6 \times 6.959 = 41.754 \)
- \( x=13 \): \( 3 \times (13 - 9.3621)^2 = 3 \times (3.6379)^2 = 3 \times 13.234 = 39.702 \)
- \( x=14 \): \( 6 \times (14 - 9.3621)^2 = 6 \times (4.6379)^2 = 6 \times 21.509 = 129.054 \)
- \( x=15 \): \( 3 \times (15 - 9.3621)^2 = 3 \times (5.6379)^2 = 3 \times 31.786 = 95.358 \)

Now, sum all these up:

143.76 + 76.112 = 219.872; +67.818 = 287.69; +33.48 = 321.17; +5.565 = 326.735; +0.524 = 327.259; +2.849 = 330.108; +13.415 = 343.523; +41.754 = 385.277; +39.702 = 424.979; +129.054 = 554.033; +95.358 = 649.391

Now, divide by \( n - 1 = 57 \):

\( s^2 = \frac{649.391}{57} \approx 11.3928 \)

Then, \( s = \sqrt{11.3928} \approx 3.375 \approx 3.38 \)

Wait, but the initial answer was 4.29, which is wrong. So the correct sample standard deviation is approximately 3.38.

Answer:

3.38