Astudy found that the mean amount of time cars spent in drive-throughs of a certain fast-food restaurant was 157.4 seconds. assuming drive-through times are normally distributed with a standard deviation of 33 seconds, complete parts below:

(a) what is the probability that a randomly selected car will get through the restaurant's drive-through in less than 105 seconds? the probability that a randomly selected car will get through the restaurant's drive-through in less than 105 seconds is (round to four decimal places as needed.)
(b) what is the probability that a randomly selected car will spend more than 195 seconds in the restaurant's drive-through? the probability that a randomly selected car will spend more than 195 seconds in the restaurant's drive-through is (round to four decimal places as needed)

The question is an illustration of a normal distribution.

The given parameters are:

(a) Probability that a selected car gets through in less than 105 seconds

This means that: x = 105

First, we calculate the z-score

This gives

So, the required probability is:

Using a z probability table, we have:

(b) Probability that a selected car spends more than 195 seconds

This means that: x = 195

First, we calculate the z-score

This gives

So, the required probability is:

Using a z probability table, we have:

Read more about probabilities of normal distributions at:

link

a) The probability that a randomly selected car will get through the restaurant's drive-through in less than 105 seconds is 5.59%.

b) The probability that a randomly selected car will spend more than 195 seconds in the restaurant's drive-through is 12.71%

Step-by-step explanation:

Normally distributed problems can be solved by the z-score formula:

On a normaly distributed set with mean and standard deviation , the z-score of a value X is given by:

After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.

A study found that the mean amount of time cars spent in drive-throughs of a certain fast-food restaurant was 157.4 seconds. This means that

Assuming drive-through times are normally distributed with a standard deviation of 33 seconds. This means that .

(a) What is the probability that a randomly selected car will get through the restaurant's drive-through in less than 105 seconds?

This probability is the pvalue of the zscore of

has a pvalue of .0559. This means that there is a 5.59% probability that  a randomly selected car will get through the restaurant's drive-through in less than 105 seconds,

(b) What is the probability that a randomly selected car will spend more than 195 seconds in the restaurant's drive-through?

This is 1 subtracted by the pvalue of the zscore of .

has a pvalue of .87286. This means that there is a 87.286% probability that  a randomly selected car will get through the restaurant's drive-through in less than 195 seconds.

So the probability that a randomly selected car will spend more than 195 seconds in the restaurant's drive through is

the answer   is 200

step-by-step explanation:

this is a problem about poisson probabilities.   i don't know if you've

already studied poisson distributions, or if you have a place to look

to read more about the subject.  

here's how you can see a poisson distribution coming.   when there are

a lot of events that may happen at any time (or any place), and when

you're counting how many of them happen at a particular time (or a

particular place), and each event is independent of the others, then

the result is a poisson distribution.   this is roughly true for the

chocolate chips; one chip being in a cookie has only a small effect on

the probability that another chip will be in the cookie.

once you know you have a poisson distribution, you know a lot.   there

is only one parameter to a poisson distribution, so if you know the

mean, you know everything.   the form of the distribution is

  probability of n chips = {[e^(-x)]*(x^n)} / (n! )