**Please show work and calculations, thanks!! Fast food restaurants spend time s

Question

**Please show work and calculations, thanks!!

Fast food restaurants spend time studying the amount of time cars spend in their drive-through. QSR Magazine studies drive through times each year. Wendy’s has the fastest drive through time with an average of 133.63 seconds in the most recent study. Answer these questions below assuming drive through times are normally distributed with a standard deviation of 28 seconds.

1). What is the probability that a randomly selected car will get through Wendy’s drive-through in less than 100 seconds? (.5 pts)

2.)What percentage of people will have spent more than 160 seconds in the drive through? (.5 pts)

3.)What percentage of cars spend between 2 and 3 minutes in the drive through? (1.5 pts)

4.) Would it be unusual for a car to spend more than 3.5 minutes in Wendy’s drive through? Why? (1 pt)

The mean incubation time of fertilized chicken eggs kept in an incubator is 21 days. Suppose that the incubation times are normally distributed and the standard deviation is 1.3 days.

1). What is the 22nd percentile for incubation times of chicken eggs? (1.5 pts)

2). Determine the two incubation times that mark off the middle (center) 95% of fertilized chicken eggs. (1.5 pts)

A major exam company retains the ACT scores for all test takers for the last year. The mean composite ACT score for last year was 21.0 with a standard deviation of 5.4.

                     1.) If were to create a sampling distribution from this data with samples of 30 exams     from the data, what would the shape, mean, and standard deviation of distribution be? Why? (1.5 pts)

2.) Would the shape, mean, or standard deviation of the sampling distribution vary if our sample size increased to 100 exams? If so, which ones would vary and why? (1.5 pts)

Explanation / Answer

Q1.

Mean ( u ) =133.63
Standard Deviation ( sd )=28
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a.
P(X < 100) = (100-133.63)/28
= -33.63/28= -1.2011
= P ( Z <-1.2011) From Standard Normal Table
= 0.1149                  
b.
P(X > 160) = (160-133.63)/28
= 26.37/28 = 0.9418
= P ( Z >0.942) From Standard Normal Table
= 0.1732                  
c.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 120) = (120-133.63)/28
= -13.63/28 = -0.4868
= P ( Z <-0.4868) From Standard Normal Table
= 0.31321
P(X < 180) = (180-133.63)/28
= 46.37/28 = 1.6561
= P ( Z <1.6561) From Standard Normal Table
= 0.95115
P(120 < X < 180) = 0.95115-0.31321 = 0.6379                  
d.
P(X > 210) = (210-133.63)/28
= 76.37/28 = 2.7275
= P ( Z >2.728) From Standard Normal Table
= 0.0032                  
it is unusual for a car to spend more than 3.5 minutes

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