Question

approximate the mean for the following data set. round your answer to one decimal place.

class frequency
0-11 6
12-23 12
24-35 4
36-47 15
48-59 11

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your answer
the mean is approximately 9.6

Explanation:

Step1: Find the midpoint of each class

For class \(0 - 11\), midpoint \(x_1=\frac{0 + 11}{2}=5.5\)
For class \(12 - 23\), midpoint \(x_2=\frac{12 + 23}{2}=17.5\)
For class \(24 - 35\), midpoint \(x_3=\frac{24 + 35}{2}=29.5\)
For class \(36 - 47\), midpoint \(x_4=\frac{36 + 47}{2}=41.5\)
For class \(48 - 59\), midpoint \(x_5=\frac{48 + 59}{2}=53.5\)

Step2: Calculate the product of midpoint and frequency for each class

For class \(0 - 11\): \(x_1\times f_1 = 5.5\times6 = 33\)
For class \(12 - 23\): \(x_2\times f_2 = 17.5\times12 = 210\)
For class \(24 - 35\): \(x_3\times f_3 = 29.5\times4 = 118\)
For class \(36 - 47\): \(x_4\times f_4 = 41.5\times15 = 622.5\)
For class \(48 - 59\): \(x_5\times f_5 = 53.5\times11 = 588.5\)

Step3: Find the sum of all frequencies

\(\sum f = 6 + 12 + 4 + 15 + 11 = 48\)

Step4: Find the sum of all \(x\times f\) products

\(\sum (x\times f)=33 + 210 + 118 + 622.5 + 588.5 = 1572\)

Step5: Calculate the mean

Mean \(\bar{x}=\frac{\sum (x\times f)}{\sum f}=\frac{1572}{48}=32.75\approx32.8\) (rounded to one decimal place)

Answer:

32.8