To estimate the mean from a histogram, you typically follow these steps:
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Calculate the Midpoint for Each Bin: Determine the midpoints of the ranges represented by each bar in the histogram. Each bin will have a range determined by its left and right edges.
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Multiply Midpoints by Frequency: For each bin, multiply the midpoint by the frequency of that bin (the height of the bar).
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Sum these Products: Sum all the products you calculated in step 2.
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Calculate Total Frequency: Add up all the frequencies to get the total number of observations.
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Calculate the Mean: Divide the sum of the products by the total frequency.
Since I cannot directly view the histogram you mentioned, here's how you would do this process:
Assuming you have the frequencies for each bin (which correspond to the ranges of 21-24, 24-27, etc.) and can determine the midpoints accordingly (for example, the midpoint of 21-24 is 22.5), you would proceed as follows:
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Estimate Midpoints:
- For the first bin (21-24), midpoint = 22.5
- For the second bin (24-27), midpoint = 25.5
- Continue for each bin.
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Frequency Values: Assume the frequencies for each bin are as follows (example values):
- 21-24: 4
- 24-27: 6
- 27-30: 8
- 30-33: 10
- 33-36: 5
- 36-39: 3
- 39-42: 2
- 42-45: 1
- 45-48: 1
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Calculate Products:
- For 21-24: \( 22.5 * 4 = 90 \)
- For 24-27: \( 25.5 * 6 = 153 \)
- For 27-30: \( 28.5 * 8 = 228 \)
- For 30-33: \( 31.5 * 10 = 315 \)
- For 33-36: \( 34.5 * 5 = 172.5 \)
- For 36-39: \( 37.5 * 3 = 112.5 \)
- For 39-42: \( 40.5 * 2 = 81 \)
- For 42-45: \( 43.5 * 1 = 43.5 \)
- For 45-48: \( 46.5 * 1 = 46.5 \)
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Sum Products: Add all the products you computed. Assume that the sum is \( 90 + 153 + 228 + 315 + 172.5 + 112.5 + 81 + 43.5 + 46.5 = 1097 \).
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Total Frequency: Add all the frequencies: \( 4 + 6 + 8 + 10 + 5 + 3 + 2 + 1 + 1 = 40 \).
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Calculate Mean: \[ \text{Mean} = \frac{\text{Sum of Products}}{\text{Total Frequency}} = \frac{1097}{40} \approx 27.425 \]
In this example, the estimated mean would be approximately \( 27.43 \).
Reasonableness: This estimate is reasonable if the majority of the frequencies are clustered around the mid-range values of the distribution. The histogram shape and location of frequencies will influence how closely the estimated mean aligns with actual mean. If the distribution is symmetrical and unimodal, such a mean will generally be an accurate representation of the data set's central tendency.