Answer:
Kindly check explanation
Step-by-step explanation:
Given the following :
Mean commute time (m) = 27.3 minutes
Standard deviation (sd) = 7.1 minutes
a) What minimum percentage of commuters in the city has a commute time within 2 standard deviations of the mean?
Using chebyshev's theorem ;
No more than 1/k² values of a distribution can be
k standard deviations from the mean.
Here k = 2
Hence,
1/k² = 1/2² = 1/4 = 0.25
Hence, minimum percentage of commmuters within 2 standard deviations of the mean :
1 - 0.25 = 0.75 m
B.) What minimum percentage of commuters in the city has a commute time within 1.5 standard deviations of the mean?
Here, k = 1.5
1/k² = 1/1.5² = 1/2.25 = 0.444
Hence, minimum commute time within 2 standard deviations of the mean :
1 - 0.444 = 0.56 = 56%
Commute time within 1.5 standard deviation
[27.3 - (1.5 × 7.1), 27.3 + (1.5 * 7.1)
[16.65, 37.95]
C.) What is the minimum percentage of commuters who have commute times between 6minutes and 48.6 minutes?
X - m /sd
X = 6
= (6 - 27.3) / 7.1 = - 3
X = 48.6
= (48.6 - 27.3) / 7. 1 = 3
Hence,
1/k² = 1 / 3² = 1/9
1 - 1/9 = 8/9
= 0.888 = 0.89 = 89%
