Question

in a distribution of 172 values with a mean of 72, at least 129 fall within the interval 68-76. approximately what percentage of values should fall in the interval 64-80? use chebyshev’s theorem. round your k to one decimal place, your s to two decimal places, and the final answer to two decimal places.
approximately □% of data will fall between 64 and 80.

Explanation:

Step1: Find the proportion for 68 - 76

First, find the proportion of values in the interval 68 - 76. The total number of values is \( n = 172 \), and the number of values in the interval is \( 129 \). So the proportion \( p_1=\frac{129}{172}\approx0.75 \).

Step2: Use Chebyshev's theorem for 68 - 76

Chebyshev's theorem states that the proportion of data within \( k \) standard deviations of the mean is at least \( 1-\frac{1}{k^2} \). For the interval 68 - 76, the mean \( \mu = 72 \), so the distance from the mean is \( 72 - 68 = 4 \) and \( 76 - 72 = 4 \). So the interval is \( \mu\pm k_1s \), where \( k_1s = 4 \). We know that the proportion is at least \( 0.75 \), so \( 1-\frac{1}{k_1^2}=0.75 \). Solving for \( k_1 \):
\[ \begin{align*} \frac{1}{k_1^2}&=1 - 0.75\\ \frac{1}{k_1^2}&=0.25\\ k_1^2&=\frac{1}{0.25}=4\\ k_1& = 2 \end{align*} \]
Since \( k_1s = 4 \) and \( k_1 = 2 \), we can find the standard deviation \( s \): \( s=\frac{4}{k_1}=\frac{4}{2}=2.00 \) (wait, no, wait. Wait, maybe I made a mistake. Wait, the interval is \( \mu\pm k_1s \), so \( \mu - k_1s=68 \) and \( \mu + k_1s = 76 \), so \( k_1s=4 \). But we found \( k_1 = 2 \) from the proportion, so \( s=\frac{4}{k_1}=\frac{4}{2}=2 \)? Wait, but let's check again. Wait, the proportion is \( 0.75 \), so \( 1-\frac{1}{k_1^2}=0.75 \implies \frac{1}{k_1^2}=0.25 \implies k_1^2 = 4 \implies k_1 = 2 \). So \( k_1s = 4 \implies s=\frac{4}{k_1}=\frac{4}{2}=2 \). Wait, but maybe I messed up the proportion. Wait, the problem says "at least 129 fall within the interval 68 - 76", so the proportion is at least \( \frac{129}{172}\approx0.75 \), so Chebyshev's theorem gives that \( 1-\frac{1}{k^2}\leq p \), but here the proportion is \( 0.75 \), so \( 1-\frac{1}{k^2}=0.75 \) gives \( k = 2 \), so that's correct. So \( k_1 = 2 \), and \( k_1s = 4 \implies s = 2 \). Wait, but let's check the next interval.

Step3: Find \( k \) for 64 - 80

For the interval 64 - 80, the mean is \( \mu = 72 \), so the distance from the mean is \( 72 - 64 = 8 \) and \( 80 - 72 = 8 \). So the interval is \( \mu\pm k_2s \), where \( k_2s = 8 \). We know \( s = 2 \) (from before? Wait, no, wait, maybe my calculation of \( s \) was wrong. Wait, wait, let's re - evaluate. Wait, the proportion \( \frac{129}{172}\approx0.75 \), so Chebyshev's theorem says that the proportion within \( k \) standard deviations is at least \( 1-\frac{1}{k^2} \). So if the proportion is \( 0.75 \), then \( 1-\frac{1}{k^2}=0.75 \implies k = 2 \), as before. Then the interval \( 68 - 76 \) is \( \mu\pm 2s \), so \( \mu - 2s=68 \) and \( \mu + 2s = 76 \). So \( 2s=4 \implies s = 2 \). Now, for the interval \( 64 - 80 \), the distance from the mean is \( 72 - 64 = 8 \) and \( 80 - 72 = 8 \), so this is \( \mu\pm k_2s \), where \( k_2s = 8 \). Since \( s = 2 \), then \( k_2=\frac{8}{s}=\frac{8}{2}=4 \)? Wait, no, wait, \( k_2s = 8 \), and \( s = 2 \), so \( k_2=\frac{8}{2}=4 \)? Wait, but let's use Chebyshev's theorem. Wait, maybe I made a mistake in calculating \( s \). Wait, let's start over.

Alternative approach:

First, proportion in 68 - 76: \( \frac{129}{172}\approx0.75 \). Chebyshev's theorem: proportion within \( k \) SDs is at least \( 1-\frac{1}{k^2} \). So \( 1-\frac{1}{k^2}=0.75 \implies k^2 = 4 \implies k = 2 \). So the interval 68 - 76 is \( \mu\pm 2s \), so \( \mu - 2s=68 \), \( \mu + 2s = 76 \). \( \mu = 72 \), so \( 72-2s = 68 \implies 2s=4 \implies s = 2 \). Now, for the interval 64 - 80: \( \mu - k_2s=64 \), \( \mu + k_2s = 80 \). So \( 72 - k_2s=64 \implies k_2s=8 \). Since \( s = 2 \), \( k_2=\frac{8}{2}=4 \)? Wait, no, \( k_2s = 8 \), \( s = 2 \), so \( k_2 = 4 \)? But Chebyshev's theorem for \( k = 4 \): the proportion is at least \( 1-\frac{1}{k^2}=1-\frac{1}{16}=\frac{15}{16}\approx0.9375 \). But that seems too high. Wait, maybe my calculation of \( s \) is wrong. Wait, maybe the first interval is \( \mu\pm k_1s \), and we need to find \( s \) from the proportion. Wait, let's do it correctly.

Let's denote:

For the first interval (68 - 76):

\( \mu = 72 \), so the lower bound is \( \mu - k_1s=68 \), upper bound is \( \mu + k_1s = 76 \). So the length from mean is \( k_1s=4 \).

The proportion of data in this interval is \( \frac{129}{172}\approx0.75 \). By Chebyshev's theorem, the proportion of data within \( k_1 \) standard deviations is at least \( 1-\frac{1}{k_1^2} \). So:

\( 1-\frac{1}{k_1^2}\leq0.75 \) (wait, no: Chebyshev says "at least", so the proportion is at least \( 1-\frac{1}{k_1^2} \), and we know that the proportion is \( 0.75 \), so \( 1-\frac{1}{k_1^2}\leq0.75 \)? No, wait, no: the proportion is at least \( 1-\frac{1}{k_1^2} \), so if the observed proportion is \( 0.75 \), then \( 1-\frac{1}{k_1^2}\leq0.75 \)? No, that's not right. Wait, Chebyshev's theorem is a lower bound. So the actual proportion is at least \( 1-\frac{1}{k_1^2} \). So if the actual proportion is \( 0.75 \), then \( 1-\frac{1}{k_1^2}\leq0.75 \)? No, that's incorrect. Wait, no: the theorem says that the proportion is at least \( 1-\frac{1}{k^2} \). So if we observe that the proportion is \( 0.75 \), then \( 1-\frac{1}{k^2}\leq0.75 \)? No, that's not. Wait, no, the theorem gives a lower bound, so the actual proportion is greater than or equal to \( 1-\frac{1}{k^2} \). So if the actual proportion is \( 0.75 \), then \( 1-\frac{1}{k^2}\leq0.75 \)? No, that's the opposite. Wait, no: \( \text{Proportion}\geq1 - \frac{1}{k^2} \). So if the proportion is \( 0.75 \), then \( 0.75\geq1-\frac{1}{k^2} \), which implies \( \frac{1}{k^2}\geq1 - 0.75 = 0.25 \), which implies \( k^2\leq4 \), so \( k\leq2 \). But we know that the interval is \( \mu\pm k_1s \), with \( k_1s = 4 \). Let's assume that the proportion is exactly \( 0.75 \) (since it's "at least" 129, but we can use the proportion to find \( k \)). So if we take the proportion as \( 0.75 \), then \( 1-\frac{1}{k^2}=0.75 \), so \( k = 2 \), as before. Then \( k_1s = 4 \), so \( s=\frac{4}{k_1}=\frac{4}{2}=2 \).

Now, for the interval 64 - 80: the distance from the mean is \( 72 - 64 = 8 \) and \( 80 - 72 = 8 \), so this is \( \mu\pm k_2s \), where \( k_2s = 8 \). Since \( s = 2 \), then \( k_2=\frac{8}{s}=\frac{8}{2}=4 \)? Wait, no, \( k_2s = 8 \), so \( k_2=\frac{8}{s} \). But we found \( s = 2 \), so \( k_2 = 4 \). Then, using Chebyshev's theorem, the proportion of data within \( k_2 \) standard deviations is at least \( 1-\frac{1}{k_2^2} \). So for \( k_2 = 4 \):
\[ 1-\frac{1}{k_2^2}=1-\frac{1}{16}=\frac{15}{16}\approx0.9375 \]
Wait, but that seems high. Wait, maybe I made a mistake in calculating \( s \). Wait, let's check the first interval again. The mean is 72, interval 68 - 76. So the difference from the mean is 4 (72 - 68 = 4, 76 - 72 = 4). So the interval is \( \mu\pm4 \). If we use Chebyshev's theorem, the proportion within \( k \) standard deviations is at least \( 1-\frac{1}{k^2} \). We know that at least 129 out of 172 are in this interval, so the proportion is at least \( \frac{129}{172}\approx0.75 \). So \( 1-\frac{1}{k^2}\leq0.75 \)? No, no, the theorem says the proportion is at least \( 1-\frac{1}{k^2} \), so \( 1-\frac{1}{k^2}\leq\text{proportion} \). So \( 1-\frac{1}{k^2}\leq0.75 \implies \frac{1}{k^2}\geq0.25 \implies k^2\leq4 \implies k\leq2 \). But the interval is \( \mu\pm4 \), so \( k s=4 \). If \( k = 2 \), then \( s = 2 \), as before.

Now, for the interval 64 - 80, the interval is \( \mu\pm8 \) (since 72 - 64 = 8, 80 - 72 = 8). So this is \( \mu\pm k_2s \), where \( k_2s = 8 \). Since \( s = 2 \), then \( k_2=\frac{8}{s}=\frac{8}{2}=4 \). Then the proportion of data within \( k_2 = 4 \) standard deviations is at least \( 1-\frac{1}{k_2^2}=1-\frac{1}{16}=\frac{15}{16}\approx0.9375 \), or 93.75%.

Wait, but let's verify again. Wait, maybe the standard deviation is not 2. Wait, maybe I messed up the first step. Let's recalculate the proportion: \( \frac{129}{172}\approx0.75 \), so that's 75%. Chebyshev's theorem: for \( k = 2 \), the proportion is at least \( 1-\frac{1}{4}=0.75 \), which matches. So that means that the interval 68 - 76 is exactly \( \mu\pm 2s \), so \( 2s = 4 \implies s = 2 \). Then the interval 64 - 80 is \( \mu\pm 4s \) (since 8 = 4*2), so \( k = 4 \). Then the proportion is at least \( 1-\frac{1}{4^2}=1-\frac{1}{16}=\frac{15}{16}\approx0.9375 \), so 93.75%.

Answer:

\( 93.75 \)