Question

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 any​outliers?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 _____.

Answer 1

Answer:

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 = $\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$,

where:

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

$\mu$ = 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 = $\mu \pm *Z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$,

Where $Z_{\alpha/2}$  ==> 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)))