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.)

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%

distance covered by pamela in 9 l of fuel = 99km

let us apply unitary method here.

distance covered by pamela in 1 l of fuel =99/9 =11 km

now let us find distance covered by pamela if she has 12 l of fuel.

distance covered in 12 l = distance covered in 1 l of fuel * 12

we know that distance covered by here in 1 l is 11 km.

so distance covered in 12 l of fuel = 11*12 = 132 km.