Question

The mean of the commute time to work for a resident of a certain city is 27.3 minutes. Assume that the standard deviation of the commute time is 7.1 minutes to complete parts​ (a) through​ (c).

​(a) What minimum percentage of commuters in the city has a commute time within 2 standard deviations of the​ mean?
nothing​% ​(Type an integer or a​ decimal.)
​(b) What minimum percentage of commuters in the city has a commute time within 1.5 standard deviations of the​ mean? What are the commute times within 1.5 standard deviations of the​ mean?
The minimum percentage of commuters in the city that has a commute time within 1.5 standard deviations of the mean is nothing​%. ​(Round to one decimal place as​ needed.)
The commute times within 1.5 standard deviations of the mean are between nothing and nothing.
​(Type an integer or a decimal. Use ascending​ order.)
​(c) What is the minimum percentage of commuters who have commute times between 6  minutes and 48.6 ​minutes?
nothing​% ​(Round to one decimal place as​ needed.)

Answer 1

Kindly check explanation

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%