The weight of a male American Bison has approximately a normal distribution with mean = 1439 pounds and a standard deviation of a 150 pounds. Suppose we take a random sample of n=81 male American bison. Let X be the random variable representing the mean weight in pounds of the 81 sampled male American bison. Let Xtot be the random variable representing sum of the weights of the 81 sampled male American bison. a) About what proportion of male American bison have weights between 1400 and 1500 pounds? b) About what proportion of male American bison have weights greater than 1500 pounds? c) About how many of the 81 sampled male American bison would you expect to have weight greater than 1500 pounds? (nearest integer)? d) About how many of the 81 sampled male American bison would you expect to have weight between 1400 and 1500 pounds? (nearest integer)? e) What is the standard deviation of the distribution of X (in pounds)? f) What is the standard deviation of the distribution of Xtot (in pounds)? g) What is the probability that 1430 < X < 1440 pounds? h) What is the probability that Xtot >118000 pounds?
The weight of a male American Bison has approximately a normal distribution with mean = 1439 pounds and a standard deviation of a 150 pounds. Suppose we take a random sample of n=81 male American bison. Let X be the random variable representing the mean weight in pounds of the 81 sampled male American bison. Let Xtot be the random variable representing sum of the weights of the 81 sampled male American bison. a) About what proportion of male American bison have weights between 1400 and 1500 pounds? b) About what proportion of male American bison have weights greater than 1500 pounds? c) About how many of the 81 sampled male American bison would you expect to have weight greater than 1500 pounds? (nearest integer)? d) About how many of the 81 sampled male American bison would you expect to have weight between 1400 and 1500 pounds? (nearest integer)? e) What is the standard deviation of the distribution of X (in pounds)? f) What is the standard deviation of the distribution of Xtot (in pounds)? g) What is the probability that 1430 < X < 1440 pounds? h) What is the probability that Xtot >118000 pounds?
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:The weight of a male American Bison has approximately a normal distribution with mean μ= 1439 pounds and a standard deviation of o= 150 pounds. Suppose we take a random sample
of n=81 male American bison. Let X be the random variable representing the mean weight in pounds of the 81 sampled male American bison. Let Xtot be the random variable
representing sum of the weights of the 81 sampled male American bison.
a) About what proportion of male American bison have weights between 1400 and 1500 pounds?
b) About what proportion of male American bison have weights greater than 1500 pounds?
c) About how many of the 81 sampled male American bison would you expect to have weight greater than 1500 pounds? (nearest integer)?
d) About how many of the 81 sampled male American bison would you expect to have weight between 1400 and 1500 pounds? (nearest integer)?
e) What is the standard deviation of the distribution of X (in pounds)?
f) What is the standard deviation of the distribution of Xtot (in pounds)?
g) What is the probability that 1430 < X < 1440 pounds ?
h) What is the probability that Xtot >118000 pounds?
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Step 1: Determine the given variables
VIEWStep 2: Find the proportion of a male bison for" between "
VIEWStep 3: Find the proportion of a male bison for " greater "
VIEWStep 4: Find no. of male bison expect to have weight for " greater " than 1500
VIEWStep 5: Find no. of male bison expect to have weight for " between "
VIEWStep 6: Find the standard deviation of sample mean
VIEWStep 7: Find the standard deviation of total
VIEWStep 8: Find proportion of sample mean for "between "
VIEWStep 9: Find the proportion of xtot is greater than 118000
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