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a purchasing manager at a university is investigating which brand of lc…

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 lumen output (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:

lifetime rangefrequency
750 to 7999
800 to 84911
850 to 8998
900 to 9495
950 to 9994

she needs to summarize the frequency distribution, using the midpoint of each data class, estimate the mean lifetime for the light bulbs in the company’s test. for your intermediate computations, use four or more decimal places, and round your answer to one decimal place.

Explanation:

Step1: Find midpoints of each class

  • For 700 - 749: Midpoint $=\frac{700 + 749}{2}=724.5$
  • For 750 - 799: Midpoint $=\frac{750 + 799}{2}=774.5$
  • For 800 - 849: Midpoint $=\frac{800 + 849}{2}=824.5$
  • For 850 - 899: Midpoint $=\frac{850 + 899}{2}=874.5$
  • For 900 - 949: Midpoint $=\frac{900 + 949}{2}=924.5$
  • For 950 - 999: Midpoint $=\frac{950 + 999}{2}=974.5$

Step2: Multiply midpoint by frequency

  • 724.5 * 5 = 3622.5
  • 774.5 * 9 = 6970.5
  • 824.5 * 11 = 9069.5
  • 874.5 * 8 = 6996
  • 924.5 * 3 = 2773.5
  • 974.5 * 2 = 1949

Step3: Sum the products

Total = 3622.5 + 6970.5 + 9069.5 + 6996 + 2773.5 + 1949
= (3622.5 + 6970.5) + (9069.5 + 6996) + (2773.5 + 1949)
= 10593 + 16065.5 + 4722.5
= 10593 + 20788
= 31381

Step4: Find total frequency

Total frequency = 5 + 9 + 11 + 8 + 3 + 2 = 38

Step5: Calculate mean

Mean $=\frac{31381}{38}\approx825.82$

Answer:

The mean number of hours is approximately 825.82.