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Math  /  Data & Statistics

QuestionWhat's the temperature? The temperature in a certain location was recorded each day for two months. The mean temperature was 21.8∘F21.8^{\circ} \mathrm{F} with a standard deviation 1.1∘F1.1^{\circ} \mathrm{F}. What can you determine about these data by using Chebyshev's Inequality with K=3K=3 ?
At least □\square %\% of the days had temperatures between □\square ∘F{ }^{\circ} \mathrm{F} and □\square ∘F{ }^{\circ} \mathrm{F}

Studdy Solution

STEP 1

1. The mean temperature μ\mu is 21.8∘F21.8^{\circ} \mathrm{F}.
2. The standard deviation σ\sigma is 1.1∘F1.1^{\circ} \mathrm{F}.
3. We are using Chebyshev's Inequality with K=3K=3 to determine the proportion of data within a certain range.

STEP 2

1. State Chebyshev's Inequality.
2. Calculate the proportion of days within K=3K=3 standard deviations of the mean.
3. Determine the temperature range within K=3K=3 standard deviations of the mean.
4. Combine the results to find the percentage and temperature range.

STEP 3

Chebyshev's Inequality states that for any random variable with finite mean and variance, at least (1−1K2)(1 - \frac{1}{K^2}) proportion of the data lies within KK standard deviations of the mean.
P(∣X−μ∣<Kσ)≥1−1K2 P(|X - \mu| < K\sigma) \geq 1 - \frac{1}{K^2}

STEP 4

Substitute K=3K=3 into Chebyshev's Inequality to find the proportion of days within 33 standard deviations of the mean.
P(∣X−21.8∣<3⋅1.1)≥1−132 P(|X - 21.8| < 3 \cdot 1.1) \geq 1 - \frac{1}{3^2}
P(∣X−21.8∣<3.3)≥1−19 P(|X - 21.8| < 3.3) \geq 1 - \frac{1}{9}
P(∣X−21.8∣<3.3)≥89 P(|X - 21.8| < 3.3) \geq \frac{8}{9}

STEP 5

Calculate the percentage of days within 33 standard deviations of the mean.
89≈0.8889 \frac{8}{9} \approx 0.8889
0.8889⋅100≈88.89% 0.8889 \cdot 100 \approx 88.89\%

STEP 6

Determine the temperature range by adding and subtracting 33 standard deviations from the mean.
21.8−3⋅1.1=21.8−3.3=18.5∘F 21.8 - 3 \cdot 1.1 = 21.8 - 3.3 = 18.5^{\circ} \mathrm{F}
21.8+3⋅1.1=21.8+3.3=25.1∘F 21.8 + 3 \cdot 1.1 = 21.8 + 3.3 = 25.1^{\circ} \mathrm{F}

STEP 7

Combine the results to find the final answer.
At least 88.89%88.89\% of the days had temperatures between 18.5∘F18.5^{\circ} \mathrm{F} and 25.1∘F25.1^{\circ} \mathrm{F}.
Solution: At least 88.89%88.89\% of the days had temperatures between 18.5∘F18.5^{\circ} \mathrm{F} and 25.1∘F25.1^{\circ} \mathrm{F}.

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