The accompanying data represent the miles per gallon of a random sample of cars with a three-cylinder, 1.0 liter engine. a. Compute the z-score corresponding to the individual who obtained 37.8 miles per gallon. Interpret this result. b. Determine the quartiles. c. Compute and interpret the interquartile range, IQR. d. Determine the lower and upper fences. Are there anyoutliers?31.5 36.0 37.8 38.5 40.1 42.234.2 36.2 38.1 38.7 40.6 42.534.7 37.3 38.2 39.5 41.4 43.435.6 37.6 38.4 39.6 41.7 49.3The z-score corresponding to the individual is and indicates that the data value is ___ standard deviation(s) the .

1). Z score = 1.464733

Mean = 38.87917

Standard deviation = 3.609405

2). Quartiles:

Q1 = 37.025

Q2 = 38.450

Q3 = 40.800

3). IQR = 3.775

4).

CI (lower fence) = 37.4351

CI (upper fence) =  40.32323

5). There is outlier in the data set. Please see attached box plot for evidence.

Step-by-step explanation:

1). By Z score, we mean:

Z = ,

where:

x-bar ==> mean(g) = 38.87917

= 37.80

standard deviation = 3.609405

sample size (n) = 24.

2). By quantile, we mean:

Q = L + (i*(n/4) - Cf)*c

Where L is the lower class limit of the quartile class

Cf is the cumulative frequency before the quartile class

c  is the class size.

3) . IQR = Q3 - Q1

4). CI = ,

Where  ==> 1.96

In order to replicate and obtain the result, please use the R code below:

g = c(31.5, 36.0, 37.8, 38.5, 40.1, 42.2,34.2, 36.2, 38.1, 38.7, 40.6, 42.5,34.7, 37.3, 38.2, 39.5,  

41.4, 43.4,35.6, 37.6, 38.4, 39.6, 41.7, 49.3)

boxplot(g)

Z = (mean(g) - 37.8)/(sd(g)/sqrt(length(g)))

mean(g)

quantile(g)

IQR(g)

CIl = mean(g) - 1.96*(sd(g)/sqrt(length(g)))

CIU = mean(g) + 1.96*(sd(g)/sqrt(length(g)))

ruth is 6 years old

step-by-step explanation:

6 + 24 = 30,   so 6/30 = .2  

if you do 1/5 it also equals .2, so that is the answer.