Answer:
The mean is approximately 102.14.
The median is approximately 94.33.
Explanation:
To find the mean and median of this frequency distribution, we'll first calculate the midpoint of each class interval and then use these midpoints to find the mean and median.
Calculate the midpoints:
Midpoint of the class "85 to 90" = (85 + 90) / 2 = 87.5
Midpoint of the class "90 to 95" = (90 + 95) / 2 = 92.5
Midpoint of the class "95 to 100" = (95 + 100) / 2 = 97.5
Midpoint of the class "100 to 105" = (100 + 105) / 2 = 102.5
Midpoint of the class "105 to 110" = (105 + 110) / 2 = 107.5
Midpoint of the class "110 to 115" = (110 + 115) / 2 = 112.5
Calculate the product of the midpoint and frequency for each class interval:
Class "85 to 90" = 87.5 * 10 = 875
Class "90 to 95" = 92.5 * 12 = 1110
Class "95 to 100" = 97.5 * 15 = 1462.5
Class "100 to 105" = 102.5 * 14 = 1435
Class "105 to 110" = 107.5 * 12 = 1290
Class "110 to 115" = 112.5 * 7 = 787.5
Calculate the total of these products:
Total = 875 + 1110 + 1462.5 + 1435 + 1290 + 787.5 = 7150
Calculate the total frequency:
Total Frequency = 10 + 12 + 15 + 14 + 12 + 7 = 70
Now, let's find the mean:
Mean = (Total) / (Total Frequency) = 7150 / 70 = 102.14 (rounded to two decimal places)
The mean is approximately 102.14.
To find the median, we'll first determine which class interval contains the median and then calculate the median within that class interval.
The cumulative frequency up to the median is (70 / 2) = 35.
Since the cumulative frequency of the first two class intervals is 10 + 12 = 22, the median falls in the "90 to 95" class interval.
Now, we can use the formula for finding the median within a class interval:
Median = L + [(n/2 - F) * w]
Where:
L is the lower limit of the median class (90).
n is the total frequency (70).
F is the cumulative frequency of the class before the median class (22).
w is the width of the class interval (5, from 90 to 95).
Median = 90 + [(35 - 22) * 5] / 15 = 90 + (13 * 5) / 15 = 90 + 65/15 = 90 + 4.33 (rounded to two decimal places)
The median is approximately 94.33.