A study found that the mean amount of time cars spent in drive-through of a certain fast-food restaurant was 136.6400 seconds. Assuming drive-through times are normally distributed with a standard deviation of 33.4000 seconds. a) What is the probability that a randomly selected car will spend more than 179 seconds in the restaurant's drive-through?
b) What proportion of cars spend between 2 and 3 minutes in the restaurant's drive-through?

a) What is the probability that a randomly selected car will spend more than 179 seconds in the restaurant's drive-through?

0.10235

b) What proportion of cars spend between 2 and 3 minutes in the restaurant's drive-through?

= 30.917% to 90.289%

= 0.30917 to 0.90289

= 0.30917 : 090289

Step-by-step explanation:

We solve for this questions using z score formula

z-score is z = (x-μ)/σ,

where

x is the raw score,

μ is the population mean

σ is the population standard deviation.

a) What is the probability that a randomly selected car will spend more than 179 seconds in the restaurant's drive-through?

x = 179 seconds μ =136.64 seconds, σ = 33.4 seconds

z = (x - μ)/σ

= (179 - 136.64)/33.4

= 1.26826

Determining the probability value from Z-Table:

P(x≤ 179) = 0.89765

P(x< 179) = P(z = 1.26826) = 0.89765

P(x>179) = 1 - P(x<179)

= 1 - 0.89765

= 0.10235

Therefore, the probability that a randomly selected car will spend more than 179 seconds in the restaurant's drive-through is 0.10235

b) What proportion of cars spend between 2 and 3 minutes in the restaurant's drive-through?

Converting the time in minutes to seconds

1 minute = 60 seconds

2 minutes = 60 seconds × 2

= 120 seconds

3 minutes = 60 seconds × 3

= 180 seconds

For 2 minutes = 120 seconds

x = 120 seconds μ =136.64 seconds, σ = 33.4 seconds

Since we are asked to find proportion, n =

z = (x - μ)/σ

= (120 - 136.64)/33.4

= -0.4982

Determining the probability value from Z-Table:

P(x≤ 120) = 0.30917

P(x = 120) = P(z = -0.4982) = 0.30917

Converting to percentage

0.30917 × 100 = 30.917 %

This means 30.917% spend 2 minutes in the restaurant drive through

For 3 minutes

x = 180 seconds μ =136.64 seconds, σ = 33.4 seconds

z = (x - μ)/σ

= (180 - 136.64)/33.4

=1.2982

Determining the probability value from Z-Table:

P(x≤ 180) = 1.2982

P(x = 180) = P(z =1.2982) = 0.90289

Converting to percentage

0.90.289 × 100 = 90.289%

This means 90.289% spend 3 minutes in the restaurant drive through

The proportion of cars that spend between 2 and 3 minutes in the restaurant's drive-through is:

= 30.917% to 90.289%

= 0.30917 to 0.90289

= 0.30917 : 090289

c

step-by-step explanation:

it isn't d, because the interquartile range is from 27-57. and the interquartile range is basically half the data, it's pretty obvious.

it isn't b, because like i said earlier, the interquartile range is from 27-57. 57-27 is 30. the real range has to be longer

not a either. although it looks like it, 3/4 of the data is not greater than 27. but if c is wrong than this should be correct

the correct answer is c, because 13 is the point in which the plot starts

step-by-step explanation:

10.50 + taxes which added up to 11.50