Question

approximating the mean of a data set given a frequency distribution
a purchasing manager at a university is investigating which brand of lcd projector to purchase for equipping classrooms. of major concern to her is the lifetime
of the light bulbs used in the projectors. one company has published the following frequency distribution from a test of the lifetimes (in hours) of 42 bulbs used
in its lcd projectors.

based on the frequency distribution, using the midpoint of each data class, estimate the mean lifetime for the light bulbs in the companys test. for your
intermediate computations, use four or more decimal places, and round your answer to one decimal place.

hours

Explanation:

Step1: Find mid - points of each class

For the class \(700 - 749\), mid - point \(x_1=\frac{700 + 749}{2}=724.5\)
For the class \(750 - 799\), mid - point \(x_2=\frac{750+799}{2}=774.5\)
For the class \(800 - 849\), mid - point \(x_3=\frac{800 + 849}{2}=824.5\)
For the class \(850 - 899\), mid - point \(x_4=\frac{850+899}{2}=874.5\)
For the class \(900 - 949\), mid - point \(x_5=\frac{900 + 949}{2}=924.5\)
For the class \(950 - 999\), mid - point \(x_6=\frac{950+999}{2}=974.5\)

Step2: Calculate the sum of \(f_ix_i\)

Given \(f_1 = 5,f_2=7,f_3 = 11,f_4=9,f_5 = 6,f_6=4\)
\(f_1x_1=5\times724.5 = 3622.5\)
\(f_2x_2=7\times774.5=5421.5\)
\(f_3x_3=11\times824.5 = 9069.5\)
\(f_4x_4=9\times874.5=7870.5\)
\(f_5x_5=6\times924.5 = 5547\)
\(f_6x_6=4\times974.5=3898\)
\(\sum_{i = 1}^{6}f_ix_i=3622.5+5421.5 + 9069.5+7870.5+5547+3898=35429\)

Step3: Calculate the sum of frequencies \(\sum_{i=1}^{6}f_i\)

\(\sum_{i = 1}^{6}f_i=5 + 7+11+9+6+4=42\)

Step4: Calculate the mean \(\bar{x}\)

\(\bar{x}=\frac{\sum_{i = 1}^{6}f_ix_i}{\sum_{i=1}^{6}f_i}=\frac{35429}{42}\approx843.5\)

Answer:

\(843.5\)