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Assume that the article reported correct information. Complete the following sta

ID: 3175367 • Letter: A

Question

Assume that the article reported correct information. Complete the following statements about the distribution of systolic blood pressure measurements for women over seventy-five.

(a) According to Chebyshev's theorem, at least _________ of the measurements lie between 119.4 mmHg and 147.0.

(b) According to Chebyshev's theorem, at least 8/9 (about 89%) of the measurements lie between __________ mmHg and _________ mmHg. (Round your answer to 1 decimal place.)

(c) Suppose that the distribution is bell-shaped. According to the empirical rule, approximately 68% of the measurements lie between _________ mmHg and ________ mmHg.

(d) Suppose that the distribution is bell-shaped. According to the empirical rule, approximately __________ of the measurements lie between 119.4 mmHg and 147.0 .

133.2

Explanation / Answer

Let X be the random variable that, systolic blood pressure measurement for women over seventy-five.

Given that, X has mean = 133.2 mmHg and standard deviation = 6.9 mmHg

Assume that the article reported correct information. Complete the following statements about the distribution of systolic blood pressure measurements for women over seventy-five.

a) Chebyshev's theorem states that

k = |(x - xbar) / s |

Let k>=1

Then the % of distirbution that lies within k SDs of the xbar is at least (1 - 1/k^2) *100.

We have given the interval is 119.4 mmHg to 147.0 mmHg.

Now first we have to find k then we can find %.

k for 119.4 and 147.0 :

k = |(119.4 - 133.2) / 6.9| = |-2| = 2

k = |(147.0 - 133.2) / 6.9 | = |2| = 2

SO here k=2 now we can use Chebyshev's theorem.

Percentage = (1 - 1/k^2)*100

= (1 - 1/2^2)*100

= 75%

According to Chebyshev's theorem, at least 75% of the measurements lie between 119.4 mmHg and 147.0.

b) Now in this part we have given percentage 8/9 (89%)

Percentage = 89% = 0.89

(1 - 1/k^2)*100 = 0.89

1 - 1/k^2 = 0.89/100

1 - 1/k^2 = 0.0089

1 - 0.0089 = 1/k^2

0.9911 = 1/k^2

k^2 = 1/0.9911 = 1.0

k = sqrt(1.01) = 1.0

lower limit = xbar - k*s = 133.2 - 1*6.9 = 126.3

upper limit = xbar + k*s = 133.2 + 1*6.9 = 140.1

According to Chebyshev's theorem, at least 8/9 (about 89%) of the measurements lie between 126.3 and 140.1.

c) Suppose that the distribution is bell-shaped.

Emperical rule states that 68% of the data lie within one standard deviation from the mean.

95% of the data lie within two standard deviation from the mean.

99.7% of the data lie within three standard deviation from the mean.

xbar - 1*s = 133.2 - 1*6.9 = 126.3

xbar + 1*s = 133.2 + 1*6.9 = 140.1

According to the empirical rule, approximately 68% of the measurements lie between 126.3 and 140.1 mmHg.

(d) Suppose that the distribution is bell-shaped.

We have to find k here.

k we can find by using same formula as in part a).

k = |(x-xbar) / s|

k for 119.4 and 147.0 are :

k = |(119.4 - 133.2)/6.9| = |-2| = 2

k = |(147.0 - 133.2) / 6.9| = |2| = 2

And by using emperical rule 95% of the data lie within two standard deviation from the mean.

According to the empirical rule, approximately 95% of the measurements lie between 119.4 mmHg and 147.0 .