Question

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

# of pairs of shoes	4	5	6	7	8	9	10	11

round all your answers to 4 decimal places where possible.
the mean is:
the median is:
the sample standard deviation is:
the first quartile is:
the third quartile is:
what percent of the respondents have at least 10 pairs of shoes?
64% of all respondents have fewer than how many pairs of shoes?

Explanation:

Step1: Calculate mean

Sum \(X \times f = 4*11 + 5*6 + 6*5 + 7*11 + 8*11 + 9*6 + 10*10 + 11*9 = 522\). Mean = \(522/69 \approx 7.5652\)

Step2: Find median position

Median position = \((69+1)/2 = 35\). Cumulative freq up to 7:33 <35 ≤44 (up to 8). Median=8

Step3: Compute sample variance

Sum \((X-\bar{X})^2 f \approx 374.9533\). Variance= \(374.9533/(69-1) \approx 5.5140\). Std dev=√5.5140≈2.3482

Step4: Determine Q1

Q1 position=17.5. 17th=5, 18th=6. Q1=(5+6)/2=5.5

Step5: Determine Q3

Q3 position=52.5. 52nd and 53rd in X=10. Q3=10

Step6: Percent with ≥10 pairs

Freq=10+9=19. Percent= \(19/69*100 \approx 27.5362\%\)

Step7: 64% fewer than X

64% of 69=44.16. Cumulative up to 8=44 <44.16. X=9

Answer:

7.5652
8
2.3482
5.5
10
27.5362
9