3.3 10 )Heights of women have a bell-shaped distribution with a mean of 162cm an

Question

3.3

10)Heights of women have a bell-shaped distribution with a mean of 162cm and a standard deviation of 8cm. Using Chebyshev's theorem, what do we know about the percentage of women with heights that are within 3standard deviations of the mean? What are the minimum and maximum heights that are within 3 standard deviations of the mean?

At least __________% of women have heights within 3 standard deviations of 162cm.

The minimum height that is within 3 standard deviations of the mean is _______

The maximum height that is within 3 standard deviations of the mean is_______

7)A certain group of test subjects had pulse rates with a mean of 73.1 beats per minute and a standard deviation of 11.1 beats per minute. Would it be unusual for one of the test subjects to have a pulse rate of f 145.3 beats per minute?

Minimum "usual" value_________beats per minute
(Type an integer or a decimal.)
Maximum "usual" value__________beats per minute
(Type an integer or a decimal.)

Is 73.1 beats per minute an unusual pulse rate?
A.No, because it is smaller than the minimum "usual" value.
B.Yes comma because it is larger than the maximum "usual" value.
C.Yes, because it is between the minimum and maximum "usual" values.
D.No, because it is larger than the maximum "usual" value.
E.Yes comma because it is smaller than the minimum "usual" value.
F.No comma because it is between the minimum and maximum "usual" values.

)Please answer both questions all parts)!

Explanation / Answer

10) Let X be the height of women. X follows Normal distribution with mean =162 and standard deviation 8cm.

Percentage of women having height within 3 standard deviation of mean. Given k=3

Using Chebyshev's theorem about (1-(1/k2)) % of women have height withing k standard deviation from mean.

==> 1 - (1/32) = 1 - 1/9 = 8/9 = 0.8889

About 88.89% percentage of women have height within 3 standard deviation from mean.

Minimum height within 3 standard deviation from mean = mean - (3 * sd) = 162- (3*8) = 138

Maximum height within 3 standard deviation from mean = mean + (3* sd) = 162 + (3*8) = 186

The minimum height that is within 3 standard deviations of the mean is 138

The maximum height that is within 3 standard deviations of the mean is 186

7) Let X be the pulse rate. X follows Normla distribution with mean = 73.1 and standard deviation, sd= 11.1

Anything beyond 3 standard deviation from mean is considered to be unusual.

Here X = 145.3 corresponding z is

z = (145.3 - 73.1) /11.1 = 6.50

Now P(Z > 6.50) = 1 - P(Z <= 6.50) = 1 - 1 = 0

So it is unusual to have pulse rate 145.3

So the minimum usual value = mean - (3*sd) = 73.1 - (3*11.1) = 39.8

Maximum usual value = mean + (3*sd) = 73.1 + (3*11.1) = 106.4

Is 73.1 beats per minute an unusual pulse rate?

B.Yes comma because it is larger than the maximum "usual" value.

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.