Question

in a distribution of 624 values with a mean of 72, at least 585 fall within the interval 65-79. approximately what percentage of values should fall in the interval 58-86? use chebyshevs 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 58 and 86.

Explanation:

Step1: Find k for 65 - 79 interval

The mean $\mu = 72$. For the interval $65 - 79$, the distance from the mean is $72 - 65 = 7$ and $79 - 72 = 7$. So the interval is $\mu \pm k_1 s = 72 \pm 7$. First, find the proportion of data in $65 - 79$: $\frac{585}{624} \approx 0.9375$. By Chebyshev's theorem, the proportion of data within $k$ standard deviations is at least $1 - \frac{1}{k^2}$. So $1 - \frac{1}{k_1^2} \leq 0.9375$ (wait, actually, Chebyshev says at least $1 - \frac{1}{k^2}$, so we have $1 - \frac{1}{k_1^2} \leq 0.9375$? No, wait: the proportion is at least $1 - \frac{1}{k^2}$, so $1 - \frac{1}{k_1^2} \leq 0.9375$ is wrong. Wait, the actual proportion is $\frac{585}{624} \approx 0.9375$, which is at least $1 - \frac{1}{k_1^2}$. So $1 - \frac{1}{k_1^2} \leq 0.9375$? No, $1 - \frac{1}{k_1^2}$ is the minimum proportion, so the actual proportion (0.9375) is greater than or equal to $1 - \frac{1}{k_1^2}$. So $1 - \frac{1}{k_1^2} \leq 0.9375$. Solving for $k_1$: $\frac{1}{k_1^2} \geq 1 - 0.9375 = 0.0625$, so $k_1^2 \leq \frac{1}{0.0625} = 16$, so $k_1 \leq 4$? Wait, no, that can't be. Wait, maybe I mixed up. Wait, the interval is $\mu \pm k s$, so $65 = 72 - k_1 s$ and $79 = 72 + k_1 s$, so the length from mean is $k_1 s = 7$. Also, the proportion of data is $\frac{585}{624} \approx 0.9375$, which is the actual proportion, and Chebyshev says that at least $1 - \frac{1}{k^2}$ of the data is within $k$ standard deviations. So $1 - \frac{1}{k_1^2} \leq 0.9375$? No, $1 - \frac{1}{k_1^2}$ is the minimum, so the actual proportion (0.9375) is greater than or equal to $1 - \frac{1}{k_1^2}$. So $1 - \frac{1}{k_1^2} \leq 0.9375$ → $\frac{1}{k_1^2} \geq 0.0625$ → $k_1^2 \leq 16$ → $k_1 \leq 4$. But also, from the interval, $k_1 s = 7$. Wait, maybe we can find $s$ first? Wait, no, maybe I made a mistake. Wait, let's do it correctly. Let's denote the proportion $p = \frac{585}{624} \approx 0.9375$. By Chebyshev, $p \geq 1 - \frac{1}{k^2}$, so $1 - \frac{1}{k^2} \leq p$. So $1 - \frac{1}{k^2} \leq 0.9375$ → $\frac{1}{k^2} \geq 0.0625$ → $k^2 \leq 16$ → $k \leq 4$. But also, the interval is $\mu \pm k s$, so $7 = k s$. Now, we need to find $s$. Wait, maybe we can find $k$ such that $1 - \frac{1}{k^2} = 0.9375$ (since the actual proportion is 0.9375, which is the minimum proportion? No, Chebyshev gives a lower bound, so the actual proportion is at least $1 - \frac{1}{k^2}$. So if the actual proportion is 0.9375, then $1 - \frac{1}{k^2} \leq 0.9375$, so $k$ can be 4, because $1 - \frac{1}{16} = 0.9375$. Ah! There we go. So $k_1 = 4$, because $1 - \frac{1}{4^2} = 1 - \frac{1}{16} = 0.9375$, which matches the proportion $\frac{585}{624} = 0.9375$. So $k_1 = 4$, and the interval is $\mu \pm 4s = 72 \pm 7$, so $4s = 7$ → $s = \frac{7}{4} = 1.75$. Wait, that's nice! So $s = 1.75$? Wait, $4s = 7$ → $s = 7/4 = 1.75$. Let's check: $72 - 4*1.75 = 72 - 7 = 65$, $72 + 4*1.75 = 72 + 7 = 79$. Perfect! So $s = 1.75$.

Step2: Find k for 58 - 86 interval

Now, the new interval is $58 - 86$. The mean is 72, so $58 = 72 - k_2 s$ and $86 = 72 + k_2 s$. So the distance from mean is $k_2 s = 72 - 58 = 14$ (and $86 - 72 = 14$). We know $s = 1.75$, so $k_2 = \frac{14}{s} = \frac{14}{1.75} = 8.0$? Wait, no, wait: $s$ was 1.75? Wait, $4s = 7$ → $s = 7/4 = 1.75$. Then $k_2 s = 14$ → $k_2 = 14 / 1.75 = 8$? Wait, that can't be. Wait, no, wait, maybe I made a mistake in Step1. Wait, let's recalculate Step1. The proportion is $\frac{585}{624} = 0.9375$. Chebyshev's theorem: the proportion of data within $k$ standard deviations is at least $1 - \frac{1}{k^2}$. So we have $1 - \frac{1}{k^2} \leq 0.9375$? No, $1 - \frac{1}{k^2}$ is the minimum proportion, so the actual proportion (0.9375) is greater than or equal to $1 - \frac{1}{k^2}$. So $1 - \frac{1}{k^2} \leq 0.9375$ → $\frac{1}{k^2} \geq 0.0625$ → $k^2 \leq 16$ → $k \leq 4$. But when $k = 4$, $1 - \frac{1}{16} = 0.9375$, which is exactly the proportion. So that means the data is exactly $1 - \frac{1}{k^2}$ within $k$ standard deviations, so $k = 4$, and the interval is $\mu \pm 4s = 72 \pm 7$, so $4s = 7$ → $s = 7/4 = 1.75$. That makes sense. Now, for the interval $58 - 86$: the lower bound is $58 = 72 - k_2 s$ → $k_2 s = 72 - 58 = 14$. The upper bound is $86 = 72 + k_2 s$ → $k_2 s = 86 - 72 = 14$. So $k_2 = \frac{14}{s} = \frac{14}{1.75} = 8$? Wait, that can't be, because Chebyshev's theorem is for any $k > 1$, but if $k = 8$, then the proportion of data within $k$ standard deviations is at least $1 - \frac{1}{8^2} = 1 - \frac{1}{64} = \frac{63}{64} \approx 0.9844$. But wait, let's check the standard deviation. Wait, maybe I messed up the standard deviation. Wait, let's re-express the first interval: $65 = 72 - k_1 s$ and $79 = 72 + k_1 s$, so the range from mean is $k_1 s = 7$. The proportion is 0.9375, which is $1 - \frac{1}{k_1^2} = 0.9375$ → $\frac{1}{k_1^2} = 0.0625$ → $k_1^2 = 16$ → $k_1 = 4$. So $4s = 7$ → $s = 7/4 = 1.75$. That's correct. Now, the second interval: $58 = 72 - k_2 s$ → $k_2 s = 14$ → $k_2 = 14 / 1.75 = 8$? Wait, that's a large $k$, but Chebyshev's theorem still applies. Now, the proportion of data within $k_2$ standard deviations is at least $1 - \frac{1}{k_2^2} = 1 - \frac{1}{64} = \frac{63}{64} \approx 0.9844$, so approximately 98.44%? Wait, but that seems high. Wait, maybe I made a mistake in the standard deviation. Wait, let's check the first interval again. The number of values is 624, and 585 are in 65-79. So 624 - 585 = 39 values outside. So 39/624 = 0.0625, which is 1/16. Ah! So the proportion outside is 1/16, which by Chebyshev is at most $\frac{1}{k^2}$. So $\frac{1}{k^2} \geq \frac{1}{16}$ → $k^2 \leq 16$ → $k \leq 4$. So the proportion outside is $\frac{1}{16}$, so the proportion inside is $1 - \frac{1}{16} = \frac{15}{16} = 0.9375$, which matches. So that means the data is exactly $\frac{1}{16}$ outside $k=4$ standard deviations, so $k=4$, and the interval is $\mu \pm 4s = 72 \pm 7$, so $4s = 7$ → $s = 1.75$. Now, for the interval $58 - 86$: the distance from the mean is $72 - 58 = 14$ and $86 - 72 = 14$, so the interval is $\mu \pm k_2 s = 72 \pm 14$. We know $s = 1.75$, so $k_2 = 14 / 1.75 = 8$. Then, by Chebyshev's theorem, the proportion of data within $k_2$ standard deviations is at least $1 - \frac{1}{k_2^2} = 1 - \frac{1}{64} = \frac{63}{64} \approx 0.984375$, so approximately 98.44%. Wait, but let's check the calculations again.

Wait, maybe I made a mistake in the standard deviation. Let's re-express:

First, find the proportion of data outside the interval 65-79: $\frac{624 - 585}{624} = \frac{39}{624} = \frac{1}{16}$. By Chebyshev's theorem, the proportion of data outside $k$ standard deviations is at most $\frac{1}{k^2}$. So $\frac{1}{k^2} \geq \frac{1}{16}$ → $k^2 \leq 16$ → $k \leq 4$. So the maximum proportion outside is $\frac{1}{k^2}$, so when $k=4$, the proportion outside is $\frac{1}{16}$, which matches. So that means the interval 65-79 is exactly $\mu \pm 4s$, so $4s = 72 - 65 = 7$ (or $79 - 72 = 7$), so $s = \frac{7}{4} = 1.75$. Correct.

Now, for the interval 58-86: the lower bound is 58, so $72 - 58 = 14$; upper bound is 86, so $86 - 72 = 14$. So the interval is $\mu \pm k_2 s$, where $k_2 s = 14$. We know $s = 1.75$, so $k_2 = \frac{14}{1.75} = 8$. Then, by Chebyshev's theorem, the proportion of data within $k_2$ standard deviations is at least $1 - \frac{1}{k_2^2} = 1 - \frac{1}{64} = \frac{63}{64} \approx 0.984375$, which is approximately 98.44%.

Wait, but let's confirm the steps:

1. Find the proportion of data outside the first interval: $\frac{624 - 585}{624} = \frac{39}{624} = \frac{1}{16}$.

2. By Chebyshev, the proportion outside $k$ standard deviations is at most $\frac{1}{k^2}$. So $\frac{1}{k^2} \geq \frac{1}{16}$ → $k \leq 4$. Since the proportion outside is exactly $\frac{1}{16}$, $k=4$ (because $\frac{1}{4^2} = \frac{1}{16}$).

3. The first interval is $\mu \pm 4s$, so $4s = 7$ (since $72 - 65 = 7$) → $s = \frac{7}{4} = 1.75$.

4. For the second interval, $\mu \pm k_2 s = 72 \pm 14$ (since $72 - 58 = 14$ and $86 - 72 = 14$). So $k_2 s = 14$ → $k_2 = \frac{14}{s} = \frac{14}{1.75} = 8$.

5. The proportion of data within $k_2$ standard deviations is at least $1 - \frac{1}{k_2^2} = 1 - \frac{1}{64} = \frac{63}{64} \approx 0.984375$, so approximately 98.44%.

Yes, that makes sense. So the percentage is approximately 98.44%.

Answer:

98.44