Math Problem Statement

Work out an estimate of the mean waiting time. Benjamin rewrites the table with 3 intervals, each 20 minutes wide, and calculates a new estimated mean. Work out the difference between the original estimate and Benjamin's new estimated mean.

Solution

Let's start by solving part (a) of the problem, where we need to estimate the mean waiting time using the provided table. The steps for estimating the mean are as follows:

Step 1: Find the midpoint for each time interval.

  • For 0<x100 < x \leq 10, the midpoint = 0+102=5\frac{0 + 10}{2} = 5
  • For 10<x2010 < x \leq 20, the midpoint = 10+202=15\frac{10 + 20}{2} = 15
  • For 20<x3020 < x \leq 30, the midpoint = 20+302=25\frac{20 + 30}{2} = 25
  • For 30<x4030 < x \leq 40, the midpoint = 30+402=35\frac{30 + 40}{2} = 35
  • For 40<x5040 < x \leq 50, the midpoint = 40+502=45\frac{40 + 50}{2} = 45
  • For 50<x6050 < x \leq 60, the midpoint = 50+602=55\frac{50 + 60}{2} = 55

Step 2: Multiply the midpoint by the corresponding frequency to get the total for each class interval.

  • 5×1=55 \times 1 = 5
  • 15×9=13515 \times 9 = 135
  • 25×11=27525 \times 11 = 275
  • 35×20=70035 \times 20 = 700
  • 45×27=121545 \times 27 = 1215
  • 55×32=176055 \times 32 = 1760

Step 3: Add up the totals and the frequencies.

  • Total = 5+135+275+700+1215+1760=40905 + 135 + 275 + 700 + 1215 + 1760 = 4090
  • Total frequency = 1+9+11+20+27+32=1001 + 9 + 11 + 20 + 27 + 32 = 100

Step 4: Estimate the mean.

Mean=Total sum of midpoint × frequencyTotal frequency=4090100=40.9\text{Mean} = \frac{\text{Total sum of midpoint × frequency}}{\text{Total frequency}} = \frac{4090}{100} = 40.9

So, the estimated mean waiting time is 40.9 minutes (to 1 decimal place).


Part (b): Benjamin's New Estimated Mean

Benjamin rewrites the intervals into 3 intervals of 20 minutes each. The new intervals are:

  • 0<x200 < x \leq 20: Combine the first two rows. Frequency = 1+9=101 + 9 = 10
  • 20<x4020 < x \leq 40: Combine the next two rows. Frequency = 11+20=3111 + 20 = 31
  • 40<x6040 < x \leq 60: Combine the last two rows. Frequency = 27+32=5927 + 32 = 59

Step 1: Midpoints for the new intervals.

  • For 0<x200 < x \leq 20, the midpoint = 0+202=10\frac{0 + 20}{2} = 10
  • For 20<x4020 < x \leq 40, the midpoint = 20+402=30\frac{20 + 40}{2} = 30
  • For 40<x6040 < x \leq 60, the midpoint = 40+602=50\frac{40 + 60}{2} = 50

Step 2: Multiply the midpoint by the corresponding frequency.

  • 10×10=10010 \times 10 = 100
  • 30×31=93030 \times 31 = 930
  • 50×59=295050 \times 59 = 2950

Step 3: Add up the totals and the frequencies.

  • Total = 100+930+2950=3980100 + 930 + 2950 = 3980
  • Total frequency = 10+31+59=10010 + 31 + 59 = 100

Step 4: New estimated mean.

New mean=3980100=39.8\text{New mean} = \frac{3980}{100} = 39.8

Step 5: Find the difference between the original and new mean estimates.

Difference=40.939.8=1.1\text{Difference} = 40.9 - 39.8 = 1.1

So, the difference between the original estimate and Benjamin's new estimate is 1.1 minutes (to 1 decimal place).


Do you have any questions or would you like further details on any step?

Here are 5 related questions you might consider:

  1. How does combining intervals affect the accuracy of the mean estimate?
  2. What would happen to the mean if we used more intervals with narrower widths?
  3. Can you use the weighted mean formula to check the calculations?
  4. How would the median waiting time differ from the mean in this case?
  5. Why is it important to round the final answers to 1 decimal place in such problems?

Tip: When combining intervals for a new estimation, the accuracy can decrease as details are lost from the original data.

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Math Problem Analysis

Mathematical Concepts

Statistics
Mean Calculation
Data Grouping

Formulas

Mean = (sum of midpoint × frequency) / total frequency

Theorems

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Suitable Grade Level

Grades 8-10