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a distribution of data has 80 data values, a mean of 27.8 and a standar…

Question

a distribution of data has 80 data values, a mean of 27.8 and a standard deviation of 3.2. what can be said about the number of data values between 21.4 and 34.2? at least data values are between 21.4 and 34.2. (type in the answer for the number of data values, not the percentage.)

Explanation:

Step1: Calculate the number of standard - deviations from the mean

First, find the number of standard - deviations $k$ for the lower and upper bounds. For the lower bound $x_1 = 21.4$, $k_1=\frac{\mu - x_1}{\sigma}=\frac{27.8 - 21.4}{3.2}=\frac{6.4}{3.2}=2$. For the upper bound $x_2 = 34.2$, $k_2=\frac{x_2-\mu}{\sigma}=\frac{34.2 - 27.8}{3.2}=\frac{6.4}{3.2}=2$.

Step2: Apply Chebyshev's theorem

Chebyshev's theorem states that for any data set and any constant $k>1$, the proportion of data within $k$ standard - deviations of the mean is at least $1-\frac{1}{k^{2}}$. Here, $k = 2$, so the proportion of data within 2 standard - deviations of the mean is at least $1-\frac{1}{2^{2}}=1 - \frac{1}{4}=\frac{3}{4}$.

Step3: Calculate the number of data values

We have a total of $n = 80$ data values. The number of data values within 2 standard - deviations of the mean is at least $n\times(1-\frac{1}{k^{2}})$. Substitute $n = 80$ and $k = 2$ into the formula: $80\times\frac{3}{4}=60$.

Answer:

60