Math Problem Statement

Find the estimated mean of the data summarized in the given frequency distribution. Compare the estimated mean to the actual mean. Low Temperature ​(circle◦​F) 40minus−44 45minus−49 50minus−54 55minus−59 60minus−64

Frequency 22 44 1212 55 22 Question content area bottom Part 1 The estimated mean of the frequency distribution is enter your response here degrees. ​(Round to the nearest tenth as​ needed.) Can we safely conclude that this is the actual mean from the data​ set? Why or why​ not? A. ​Yes, because mean is not affected by outliers. B. ​No, because there are one or more outliers. C. ​No, because we do not know the actual data values. We can only use the class midpoints. D. ​Yes, because the esimated mean from a frequency distribution and the actual mean from a data set are always identical.

Solution

To find the estimated mean from a frequency distribution, we use the following formula:

μ^=(f×x)f\hat{\mu} = \frac{\sum (f \times x)}{\sum f}

Where:

  • ff is the frequency of each class.
  • xx is the midpoint of each class.

Step 1: Find the midpoints of each class.

  • For the class 40−44: midpoint x=40+442=42x = \frac{40 + 44}{2} = 42
  • For the class 45−49: midpoint x=45+492=47x = \frac{45 + 49}{2} = 47
  • For the class 50−54: midpoint x=50+542=52x = \frac{50 + 54}{2} = 52
  • For the class 55−59: midpoint x=55+592=57x = \frac{55 + 59}{2} = 57
  • For the class 60−64: midpoint x=60+642=62x = \frac{60 + 64}{2} = 62

Step 2: Multiply the frequency by the corresponding midpoint.

Class (°F)Midpoint xxFrequency fff×xf \times x
40−44422222×42=92422 \times 42 = 924
45−49474444×47=206844 \times 47 = 2068
50−54521212×52=62412 \times 52 = 624
55−595755×57=2855 \times 57 = 285
60−646222×62=1242 \times 62 = 124

Step 3: Sum the f×xf \times x and the frequencies.

(f×x)=924+2068+624+285+124=4025\sum (f \times x) = 924 + 2068 + 624 + 285 + 124 = 4025 f=22+44+12+5+2=85\sum f = 22 + 44 + 12 + 5 + 2 = 85

Step 4: Calculate the estimated mean.

μ^=40258547.4°F\hat{\mu} = \frac{4025}{85} \approx 47.4 \, \text{°F}

Step 5: Interpretation of the Mean

The estimated mean is 47.4°F (rounded to the nearest tenth).

Part 2: Can we safely conclude that this is the actual mean?

The correct answer is C:

  • No, because we do not know the actual data values. We can only use the class midpoints.

The estimated mean from a frequency distribution is based on class midpoints, not the exact data values. The actual mean could be slightly different due to the spread within each class.


Would you like more details or have any questions? Here are 5 related questions you could explore:

  1. What is the formula for calculating the median in a frequency distribution?
  2. How does skewness affect the relationship between mean and median?
  3. Can the mode be estimated from a frequency distribution?
  4. What is the significance of using midpoints in a grouped frequency distribution?
  5. How would the estimated mean change if the class intervals were narrower?

Tip: When working with grouped data, always remember that the estimated mean is an approximation and may not reflect the exact mean of the original dataset.

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Math Problem Analysis

Mathematical Concepts

Statistics
Frequency Distribution
Mean Calculation

Formulas

Estimated Mean Formula: μ = Σ(f × x) / Σf
Class Midpoint Formula: midpoint = (lower bound + upper bound) / 2

Theorems

None, this problem focuses on statistical computation.

Suitable Grade Level

Grades 9-12