Mathematics, 29.02.2020 01:15 kazmiere2005oye9tz

Consider a sample with a mean of 30 and a standard deviation of 5. Use Chebyshev's theorem to determine the minimum percentage of the data within each of the following ranges. (Round your answers to the nearest integer.) (a) 20 to 40 % (b) 15 to 45 % (c) 23 to 37 % (d) 17 to 43 % (e) 13 to 47

Answers: 2

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Answers

a) We are within 2 deviations from the mean since 30 -2*5 = 20 and 30 + 2*5 = 40. So our value for k = 2 and we can find the % like this:

$\% = (1- \frac{1}{2^2}) *100 = 75\%$

b) We are within 3 deviations from the mean since 30 -3*5 = 20 and 30 + 3*5 = 45. So our value for k = 3 and we can find the % like this:

$\% = (1- \frac{1}{3^2}) *100 = 88.88\% \approx 89\%$

c) We are within 1.4 deviations from the mean since 30 -1.4*5 = 23 and 30 + 1.4*5 = 37. So our value for k = 1.4 and we can find the % like this:

$\% = (1- \frac{1}{1.4^2}) *100 = 48.97\% \approx 49\%$

d) We are within 2.6 deviations from the mean since 30 -2.6*5 = 17 and 30 + 2.6*5 = 43. So our value for k = 2.6 and we can find the % like this:

$\% = (1- \frac{1}{2.6^2}) *100 = 85.2\% \approx 85\%$

e) We are within 3.4 deviations from the mean since 30 -3.4*5 =13 and 30 + 3.4*5 = 47. So our value for k = 3.4 and we can find the % like this:

$\% = (1- \frac{1}{3.4^2}) *100 = 91.35\% \approx 91\%$

Step-by-step explanation:

For this case we need to remember what says the Chebysev theorem, and it says that we have at least $1-\frac{1}{k^2}$ of the data liying within k deviations from the mean on the interval $\bar X \pm k s$

We know that $\bar X= 30, s=5$

Part a

For this case we want the values between 20 and 40

So we are within 2 deviations from the mean since 30 -2*5 = 20 and 30 + 2*5 = 40. So our value for k = 2 and we can find the % like this:

$\% = (1- \frac{1}{2^2}) *100 = 75\%$

Part b

For this case we want the values between 15 and 45

So we are within 3 deviations from the mean since 30 -3*5 = 20 and 30 + 3*5 = 45. So our value for k = 3 and we can find the % like this:

$\% = (1- \frac{1}{3^2}) *100 = 88.88\% \approx 89\%$

Part c

For this case we want the values between 23 and 37

So we are within 1.4 deviations from the mean since 30 -1.4*5 = 23 and 30 + 1.4*5 = 37. So our value for k = 1.4 and we can find the % like this:

$\% = (1- \frac{1}{1.4^2}) *100 = 48.97\% \approx 49\%$

Part d

For this case we want the values between 17 and 43

So we are within 2.6 deviations from the mean since 30 -2.6*5 = 17 and 30 + 2.6*5 = 43. So our value for k = 2.6 and we can find the % like this:

$\% = (1- \frac{1}{2.6^2}) *100 = 85.2\% \approx 85\%$

Part e

For this case we want the values between 13 and 47

So we are within 3.4 deviations from the mean since 30 -3.4*5 =13 and 30 + 3.4*5 = 47. So our value for k = 3.4 and we can find the % like this:

$\% = (1- \frac{1}{3.4^2}) *100 = 91.35\% \approx 91\%$

Answer from: Quest

maximum area = xy = (16)(32) = 512 ft2

to maximize the area of the garden, the side of the fence parallel to the side of the house should be 32 ft long, and the other two sides should both have length 16 ft.'

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Answer from: Quest

step-by-step explanation:

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Consider a sample with a mean of 30 and a standard deviation of 5. Use Chebyshev's theorem to determ...