Homework Answers
a) Box-and-whisker plot: plot 1st
b)
5 point summary:
Minimum = 31
First quartile, Q1 = 40
Median =45
Third quartile, Q3 = 52.5
Maximum = 68
Interquartile range, IQR = Q3 - Q1 = 52.5 - 40 = 12.5
c)
| Class interval | Frequency |
| 31 - 38 | 11 |
| 39 - 46 | 24 |
| 47 - 54 | 15 |
| 55 - 62 | 7 |
| 63 - 70 | 3 |
| Total | 60 |
| Class interval | Midpoint, x | Frequency, f | fx | fx² |
| 31 - 38 | 34.5 | 11 | 379.5 | 13092.75 |
| 39 - 46 | 42.5 | 24 | 1020 | 43350 |
| 47 - 54 | 50.5 | 15 | 757.5 | 38253.75 |
| 55 - 62 | 58.5 | 7 | 409.5 | 23955.75 |
| 63 - 70 | 66.5 | 3 | 199.5 | 13266.75 |
| Total | 60 | 2766 | 131919 |
Mean, x̅ = ∑fx / n = 2766/60 = 46.1
Standard deviation, s = √((∑fx² - (∑fx)²/n)/(n-1)) = √((131919 - 2766²/60)/(60- 1))
= 8.6420 = 8.64
----
Chebyshev's theorem states that the proportion or percentage of any data set that lies within k standard deviation of the mean where k is any positive integer greater than 1 is at least 1 – 1/k^2.
0.75 = 1-1/k²
1/k² = 1 - 0.75
k² = 1/0.25
k = 2
75% Chebyshev interval centered about the mean.
Lower limit = x̅ - 2*s = 46.1 - 2*8.64 = 28.82
Upper imit = x̅ + 2*s = 46.1 + 2*8.64 = 63.38
d)
∑x = 2769
∑x² = 132179
n = 60
Mean, x̅ = Ʃx/n = 2769/60 = 46.15
Standard deviation, s = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(132179-(2769)²/60)/(60-1)]
= 8.6256 = 8.63
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