In: Statistics and Probability
3. Consider the quality of cars, as measured by the number of cars requiring extra work after assembly, in each day’s production for 15 days:
30, 34, 9, 14, 28, 9, 23, 0, 5, 23, 25, 7, 0, 3, 24
a. Find the average number of defects per day.
b. Find the median number of defects per day.
c. Find the mode number of defects per day.
d. Find the quartiles.
e. Find the extremes.
The first task is to compute the median and the quartiles. And, in order to compute the median and the quartiles, the data needs to be put into ascending order, as shown in the table below
Position | X (Asc. Order) |
1 | 0 |
2 | 0 |
3 | 3 |
4 | 5 |
5 | 7 |
6 | 9 |
7 | 9 |
8 | 14 |
9 | 23 |
10 | 23 |
11 | 24 |
12 | 25 |
13 | 28 |
14 | 30 |
15 | 34 |
Since the sample size n = 15 is odd, we have that (n+1)/2 = (15+1)/2 = 8 is an integer value, the median is computed directly by finding the value located at position 8th, which is
median = 14
Quartiles
The quartiles are computed using the table with the data in ascending order
For Q_1 we have to compute the following position:
Since 44 is an integer number, Q_1 is computed by simply locating the value that is the the 4th position in the table with data in ascending order, which means that in this case
Q_1 = 7
For Q_3 we have to compute the following position:
Since (12\) is an integer number, Q_3 is computed by locating the value that is the the 12th position in the table with data in ascending order, which means that in this case
Q_3 = 28
The interquartile range is therefore
IQR = Q_3 - Q_1 = 28 - 7 = 21
Now, we can compute the lower and upper limits for outliers:
and then, an outcome X is an outlier if X < -24.5 or if X > 59.5.
In this case since all the outcomes X are within the values of Lower = -24.5 and Upper = 59.5
Based on the data shown above, we can construct the following table with the frequency (count) of each of the values in the sample:
X | Frequency (f) |
0 | 2 |
3 | 1 |
5 | 1 |
7 | 1 |
9 | 2 |
14 | 1 |
23 | 2 |
24 | 1 |
25 | 1 |
28 | 1 |
30 | 1 |
34 | 1 |
Therefore, based on the information obtained about the number of times each value appears in the sample, it is concluded that there are multiple modes, which are the values 0, 9 and 23, that appear the same amount of times and are more frequent than the rest of the values.
The sample size is n = 15 . The provided sample data along with the data required to compute the sample mean
X | X2 | |
30 | 900 | |
34 | 1156 | |
9 | 81 | |
14 | 196 | |
28 | 784 | |
9 | 81 | |
23 | 529 | |
0 | 0 | |
5 | 25 | |
23 | 529 | |
25 | 625 | |
7 | 49 | |
0 | 0 | |
3 | 9 | |
24 | 576 | |
Sum = | 234 | 5540 |
The sample mean is computed as follows: