Question

a purchasing manager at a university is investigating which brand of lcd projector to purchase to equip \smart\ classrooms. of major concern to her is the lifetime of the light bulbs used in the projectors. one company has published the following information regarding the lifetimes of a sample of 42 bulbs used in its lcd projectors. histogram with bulb lifetime (in hours) classes: 700 - 749 (frequency 6), 750 - 799 (frequency 7), 800 - 849 (frequency 11), 850 - 899 (frequency 9), 900 - 949 (frequency 5), 950 - 999 (frequency 4) based on the histogram, using the midpoint of each data class, estimate the mean lifetime for the sample of bulbs by this company. carry your intermediate computations to at least four decimal places, and round your answer to one decimal place.

Explanation:

Step1: Find midpoints of each class

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

Step2: Multiply midpoints by frequencies

\( f_1 = 6 \), so \( x_1f_1=724.5\times6 = 4347 \)
\( f_2 = 7 \), so \( x_2f_2=774.5\times7 = 5421.5 \)
\( f_3 = 11 \), so \( x_3f_3=824.5\times11 = 9069.5 \)
\( f_4 = 9 \), so \( x_4f_4=874.5\times9 = 7870.5 \)
\( f_5 = 5 \), so \( x_5f_5=924.5\times5 = 4622.5 \)
\( f_6 = 4 \), so \( x_6f_6=974.5\times4 = 3898 \)

Step3: Sum of \( x_if_i \) and sum of frequencies

Sum of \( x_if_i \): \( 4347+5421.5 + 9069.5+7870.5+4622.5+3898=35229 \)
Sum of frequencies \( n=6 + 7+11+9+5+4 = 42 \)

Step4: Calculate mean

Mean \( \bar{x}=\frac{\sum x_if_i}{n}=\frac{35229}{42}\approx838.7857 \approx 838.8 \) (rounded to one decimal place)

Answer:

\( 838.8 \)