MDM 4U

CHAPTER THREE Answer Section

1. 5

2.

Waiting Time (s) / 0.5–5.5 / 5.5–10.5 / 10.5–15.5 / 15.5–20.5 / 20.5–25.5
Frequency / 1 / 2 / 5 / 4 / 3
Cumulative
Frequency / 1 / 3 / 8 / 12 / 15
  1. Mound-shaped distributions have decreased frequencies on either side of the interval with greatest frequency. So, there would be few people much younger or older than 18 at the concert.

4. 30

5.The mean and median need the data to have properties of numbers to be evaluated, for example, the capacity to be added and divided, and order. The mode can be evaluated with any type of data since only frequency is required.

6.No. Examples may vary, but one example is: 2, 7, 8 median = 7.

7.Arrange the data from smallest to largest, then find the median, Q2. Find the median of the data below Q2. This gives Q1. Find the median of the data above Q2. This gives Q3. The interquartile range is Q3 Q1.

8. 12

9. If all of the original data were equal the standard deviation will remain zero.

10.81.5%.

11.You would expect to find neither. Normal curves are symmetrical about the mean and since 19 and 21 are equidistant from the mean of 20, there should be the same number of each sort of wing in the basket.

12.The set in b) contains a greater percent of the data.

13.This interval includes 86% of the batteries so 14% or 0.14(1500) = 210 would be rejected.

14.The z-scores corresponding to the 90th and 93rd percentiles are 1.28 and 1.48 respectively.

Solving this system for gives

15.Alan’s z-score is smaller. Therefore, he had the more disappointing quiz mark.

16. 3.5

  1. 230

18. Almost all of the data in a normal distribution are within 3 S.D. of the mean.

19.The first increase is The second increase is

Therefore, the first increase is greater.