Question 3 (1.1 points) Suppose that diastolic blood pressures of adult women in

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Question 3 (1.1 points) Suppose that diastolic blood pressures of adult women in the United States are approximately normally distributed with mean 80.5 and standard deviation 9.9. What proportion of women have blood pressure greater than 75.9 Write only a number as your answer. Round to 4 decimal places (for example 0.0048). Do not write as a percentage. Your Answer: Answer Save Question 4 (1.1 points) The mean length of six-year-old rainbow trout in the Arolik River in Alaska is 481 millimeters with a standard deviation of 41 millimeters. Assume these lengths are normally distributed. What proportion of six-year-old rainbow trout are less than 510 millimeters long? Wnite only a number as your answer. Round to 4 decimal places (for example 0.0048). Do not write as a percentage. Your Answer:

Explanation / Answer

3.
NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 80.5
standard Deviation ( sd )= 9.9
proportion of women have blood pressure greater than 75.9
P(X > 75.9) = (75.9-80.5)/9.9
= -4.6/9.9 = -0.4646
= P ( Z >-0.4646) From Standard Normal Table
= 0.6789

4.
NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 481
standard Deviation ( sd )= 41

proportion of six year old rainbow trout are less than 510mm long
P(X < 510) = (510-481)/41
= 29/41= 0.7073
= P ( Z <0.7073) From Standard Normal Table
= 0.7603

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