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1. The duration of 40 phone calls (in minutes) for technical support is given below 12 3.3 0.5 48.7 16.7 1.2 14.8 8.2 9 5.7 11.5 17.7 3.2 20.8 7.3 8.0 0.2 51.2 3.3 5.2 12.3 24.5 13.3 7.7 13.5 4.3 13.7 10.7 18.8 15.7 3.2 28.7 16.2 23.3 9.7 4.7 6.5 0.5 45.1 5.3 Complete Table 3.5, the frequency distribution table for the call duration data Duration in Minutes 0-6 6-12 12-18 18-24 24-30 30-36 36-42 42-48 48-54 Frequency Percent a. What percentage of phone calls lasted less than 12 minutes b. What percentage of calls lasted a half hour or more? c. Represent the frequency distribution with a histogram. Use a percent scale on the vertical axis. d. Describe the shape of the distribution. Are there any gaps in the data? Outliers?
math   Statistics-And-Probability   
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1. If we sort the data, it will be easier to find the frequencies 4.7 9.7 12 12.3 12.3 13.3 13.5 13.7 14.8 15.7 4.3 0.2 5.2 0.5 5.3 0.5 5.7 1.2 6.5 3.2 7.3 3.2 7.7 3.3 8 3.3 8.2 10.7 11.5 16.2 16.717.718.8 20.8 24.5 28.7 45.1 48. 51.2 13 observations are in 0- 6; 10 are in 6-12 (consider 12 observations at class limit in 6-12) and so on... Duration in Minutes Frequency Percent 13 10 10 32.5 25 25 0-6 6-12 12-18 18-24 24-30 30-36 36-42 42-48 48-54 0 0 1 0 0 2.5 40 100 Total Percentage are calculated by dividing each class frequency by total of frequency i.e. 40 multiplied by 100, e.g. for first class100 32.5 40 a. Classes less than 12 are 0-6 and 6-12 with percentages 32.5 and 25; so 32.5 + 25-57.5% of calls lasted less than 12 minutes b. Classes more than half hour i.e. 30 minutes are 30-36, 36-42, 42-48 and 48-54, so 5 + 0 + 0 + 2.5 + 5-12.5% calls lasted half hour or more c. The histogram of frequency distribution with percentages will look likePercent call duration 35 30 25 20 15 10 0 12 18 24 30 36 42 48 54 d. From the histogram, most of the data is appears on lower side of the plot, so the data would be positively skewed. There is a data gap 30 to 42. It seems there are outliers 42 to 54 since most of the data appears to be below 3.

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