Question

the director of a community sports camp records the daily attendance. the histogram shows the data.
where does this distribution have a peak?
80 to 89 campers
which statement best describes the distribution?
the distribution is skewed left.
the distribution is symmetric.
the distribution is skewed right.

Explanation:

Peak Identification

A peak in a histogram is the bar with the highest frequency. Looking at the histogram, the bar for the interval \(80 - 89\) campers has the highest frequency (10).

Skewness Determination

• Skewed - left: The tail of the distribution is on the left - side. In a skewed - left distribution, the left - tail is longer.
• Symmetric: The left and right sides of the distribution are mirror - images.
• Skewed - right: The tail of the distribution is on the right - side. In a skewed - right distribution, the right - tail is longer.

For the given histogram, the frequencies increase from the left (lower attendance intervals: \(50 - 59\), \(60 - 69\), \(70 - 79\)) to the interval \(80 - 89\) (the peak). There is no long tail on the right. If we consider the "tail" concept (the part of the distribution with lower frequencies after the peak), there is no tail on the right. But if we assume a wrong - perception, we note that in a skewed - left distribution, the mean is less than the median. The "pull" of the lower values (if we consider the left - side as having relatively lower frequencies compared to the right - side of the peak in a wrong - sense, but actually, for a correct skewness definition:
The frequencies for \(50 - 59\) (3), \(60 - 69\) (5), \(70 - 79\) (6), \(80 - 89\) (10). If we assume the "tail" as the part with lower frequencies. The left - side (before the peak \(80 - 89\)) has a sort of "tail" when we compare to a symmetric distribution. Mathematically, for skewness, we can also think of the following:
Let \(x_1,x_2,x_3,x_4\) be the mid - points of the intervals \(50 - 59\) (\(x_1 = 54.5\)), \(60 - 69\) (\(x_2=64.5\)), \(70 - 79\) (\(x_3 = 74.5\)), \(80 - 89\) (\(x_4=84.5\)) and \(f_1 = 3\), \(f_2=5\), \(f_3 = 6\), \(f_4 = 10\).
The mean \(\bar{x}=\frac{\sum_{i = 1}^{4}f_ix_i}{\sum_{i=1}^{4}f_i}=\frac{3\times54.5 + 5\times64.5+6\times74.5 + 10\times84.5}{3 + 5+6 + 10}=\frac{163.5+322.5 + 447+845}{24}=\frac{1778}{24}\approx74.08\)
The median: Since \(n=\sum_{i = 1}^{4}f_i=24\) (even), the median is the average of the \(12^{th}\) and \(13^{th}\) values. Cumulative frequencies: \(F_1 = 3\), \(F_2=3 + 5=8\), \(F_3=8 + 6=14\). The \(12^{th}\) and \(13^{th}\) values fall in the \(70 - 79\) interval. The median \(M\) (using the formula for grouped data \(M = L+\frac{\frac{n}{2}-F}{f}\times w\), where \(L = 70\), \(n = 24\), \(F = 8\), \(f = 6\), \(w = 10\)) \(M=70+\frac{12 - 8}{6}\times10=70+\frac{40}{6}\approx76.67\)
Since the mean (\(\approx74.08\)) is less than the median (\(\approx76.67\)), the distribution is skewed left.

Answer:

• Peak: \(80\) to \(89\) campers.
• Distribution description: The distribution is skewed left.