STAT213 Tutorial sheet #1 Solutions

 

1)      We look in the cumulative relative frequency column for 30-40 to find the percentage who are under 40 which is 70.24%

2)      P(at least 20) = 100% - P(Under 20) = 100% - 3.4% = 96.6%

3)      There are 744 clients between the ages of 20 and 30 and 1087 clients between the ages of 30 and 40. So, there are 1831 clients between the ages of 20 and 40. Since there are 2739 clients in total, the percentage between 20 and 40 is 1831/2739 = 66.85%.

4)      Since the median is the 50th percentile, we look to see in which the class the cumulative relative frequency is more than 50%. We see that 30.56% of the clients are under the age of 30, but 70.24% are under the age of 40. That means the median is between the ages of 30 and 40. So we would find the median in the 30-40 age class.

5)      Since the number of times a person goes to the movies is a counting number, the data is discrete.

6)      Here is the table:

Class

Frequency

Relative Frequency

Cumulative Relative Frequency

1 to 2

5

25.00%

25.00%

3 to 4

9

45.00%

70.00%

5 to 6

4

20.00%

90.00%

7 to 8

2

10.00%

100.00%

7)      The percentage who go less than 5 times per month is the same as the percentage who go no more than 4 times per month which is 70% from the cumulative percent column.

8)      From the cumulative percent column, we see that 25% go no more than twice per month. Then, the percentage who go more than twice per month is 100% - 25% = 75%.

9)      There are 2 approaches. In the first approach, 9/20 go 3 to 4 times per month and 4/20 go 5 to 6 times per month for a total of 13/20 = 65% who go at least 3 but no more than 6 times per month. In the second approach, P(at least 3 but no more than 6) = P(no more than 6) – P(no more than 2) = 90% - 25% = 65%.

10)  You should also be able to get the mean of 3.85 and standard deviation of 1.9 on your calculator.

11)  First we need Q1 and Q3 to determine if there are any outliers. We have n=20. The position of Q1 is 20(0.25) = 5. Since this is an integer, we take the average of the 5th and 6th values which are 2 and 3 respectively. So Q1=2.5. The position of Q3 is 20(0.75) = 15. Since this is an integer, we take the average of the 15th and 16th values which are both 5. So Q3=5. The IQR = 5 – 2.5 = 2.5. So, the lower inner fence = 2.5 – 1.5(2.5) = 2.5 – 3.75 = -1.25. Since the lowest value is 1, there are no outliers on the lower end. The upper inner fence = 5 + 1.5(2.5) = 5 + 3.75 = 8.75. Since the highest value is 8, there are no outliers on the upper end. Finally, we need the median (AKA Q2). The position of Q2 is 20(0.5) = 10. Since this is an integer, we take the average of the 10th and 11th values which are both 4. So Q2=4. When we draw our box, the left corner is at Q1=2.5, the right corner is at Q3=5, and the middle line at Q2=4. The left whisker is drawn to the smallest value inside the inner fence which is 1. The right whisker is drawn to the largest value inside the inner fence which is 8.

12)   The coefficient of skewness = 3(3.85 – 4)/1.9 = -0.2369. Since this is a negative value between –1 and 0, the data is slightly skewed left.