Question

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 43 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.

bulb lifetime (in hours)	frequency
750 to 799	9
800 to 849	11
850 to 899	9
900 to 949	5
950 to 999	3

Explanation:

Step1: Find mid - points of each class

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

Step2: Calculate the product of mid - point and frequency for each class

$f_1x_1=6\times724.5 = 4347$; $f_2x_2=9\times774.5=6970.5$; $f_3x_3=11\times824.5 = 9069.5$; $f_4x_4=9\times874.5=7870.5$; $f_5x_5=5\times924.5 = 4622.5$; $f_6x_6=3\times974.5=2923.5$.

Step3: Sum up the products

$\sum_{i = 1}^{6}f_ix_i=4347+6970.5+9069.5+7870.5+4622.5+2923.5=35803.5$.

Step4: Sum up the frequencies

$\sum_{i=1}^{6}f_i=6 + 9+11+9+5+3=43$.

Step5: Calculate the mean

The mean $\bar{x}=\frac{\sum_{i = 1}^{6}f_ix_i}{\sum_{i=1}^{6}f_i}=\frac{35803.5}{43}\approx832.6$.

Answer:

$832.6$