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The histogram represents the distribution of lengths, in inches, of 25 catfish caught in a lake.

1. If possible, find the mean. If not possible, explain why not.

2. If possible, find the median. If not possible, explain why not.

3. Were any of the fish caught 12 inches long?

4. Were any of the fish caught 19 inches long?

The Histogram Represents The Distribution Of Lengths In Inches Of 25 Catfish Caught In A Lake 1 If Possible Find The Mean If Not Possible Explain Why Not 2 If P class=

Sagot :

An histogram provides a good visualization of data, however, the values of the data points are left out in an histogram

The correct responses are as follows;

1. The possible mean is given as follows;

Mean = ∑(fx)/(∑f)

Using the central values, we have;

Estimate of the Mean = (4.5×4 + 7.5×9 + 10.5×5 + 13.5×6 + 16.5×1)/(1 + 9+ 5+ 6+ 1) ≈ 10.7

2. The median is the middle value class

The total area of the histogram is given as follows

3 × (4 + 9 + 5 + 6 + 1) = 75

The middle value is located at half the total area = 75/2 = 37.5

The median class is the class of catfish with length 6 - 9 inches

3. There where 5 fishes caught that have lengths between 9 and 12 inches and there where 6 fishes caught that are 12 to 15 inches long

4. No, the longest fish caught was 18 inches long

Learn more about histograms here;

https://brainly.com/question/18406375

A histogram enables a better data visualization, but that the values of the datasets in the histogram have been left out.

Following are the responses to the given points:

  • The potential mean is provided as follows;

        Mean [tex]\bold{= \frac{\Sigma(fx)}{(\Sigma f)}}[/tex]

        We use the central values;

        Estimate of the Mean:

                   [tex]\bold{= \frac{(4.5\times 4 + 7.5 \times 9 + 10.5\times 5 + 13.5\times 6 + 16.5 \times 1)}{(1 + 9+ 5+ 6+ 1)}} \\\\ \bold{= \frac{18 + 67.5 + 52.5 + 81 + 16.5}{22}} \\\\ \bold{= \frac{235.5}{22}} \\\\ =\bold{10.704 \approx 10.7}[/tex]

  • The median is the middle class

        The entire histogram area is shown as follows:

          [tex]\to \bold{3 \times (4 + 9 + 5 + 6 + 1)= 3 \times (25)= 75}[/tex]

          This same average value is half the entire area [tex]\bold{= \frac{75}{2} = 37.5}[/tex]

          The average class seems to be catfish 6 - 9 inches long.

  • There have been 5 fish caught between 9 and 12 inches long, there have been 6 caught fish between 12 and 15 inches long.
  • No, 18 inches have been the longest caught fish.

Learn more:  

brainly.com/question/15621464

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