In: Statistics and Probability
Problem 1. A set of 25 measurements consists of the values
3 6 7 15 12
6 8 4 5 6
5 12 1 3 3
7 5 10 8 3
9 2 6 1 5
(a) Construct a relative frequency histogram to describe the data.
(10 pts) (b) Find: the
median m (5 pts), the sample mean ¯x (5 pts), the sample standard
deviation s (5 pts), the
upper and lower quartiles (10 pts), the 37-th percentile (5 pts).
(c) Apply Tchebisheff’s
Theorem in order to estimate the proportion of measurements that
lie in the interval
[¯x − (1.5)s, ¯x + (1.5)s] (10 pts).
x | frequency(f) | relative frequency(rf) | cumulative frequency(c.f.) | fx | |
---|---|---|---|---|---|
1 | 2 | 2/25 | 2 | 2 | 51.6128 |
2 | 1 | 1/25 | 3 | 2 | 16.6464 |
3 | 4 | 4/25 | 7 | 12 | 37.9456 |
4 | 1 | 1/25 | 8 | 4 | 4.3264 |
5 | 4 | 4/25 | 12 | 20 | 4.6656 |
6 | 4 | 4/25 | 16 | 24 | 2.56 |
7 | 2 | 2/25 | 18 | 14 | 1.6928 |
8 | 2 | 2/25 | 20 | 16 | 7.3728 |
9 | 1 | 1/25 | 21 | 9 | 8.5264 |
10 | 1 | 1/25 | 22 | 10 | 15.3664 |
12 | 2 | 2/25 | 24 | 24 | 70.0928 |
15 | 1 | 1/25 | 25 | 15 | 79.5664 |
a) Histogram
b) N=25
N/2=12.5
c.f. greater than 12.5 is 16 corresponding to x=6. Thus, median is 6.
N/4=6.25 ; cf greater than 6.25 is 7 which corresponds to x=3. Thus, lower quartile(Q1) = 3
3N/4=18.75; cf greater than 18.75 is 20 which corresponds to x=8. Thus, upper quartile(Q3) = 8
37N/100=9.25; cf greater than 9.25 is 12 which corresponds to x=5. Thus, 37-th percentile(P37) = 5
c) Tchebisheff's theorem is
Thus, k=1.5
Substituting these values in tchebisheff's inequality, we get