The accompanying data represent the miles per gallon of a random sample of cars with a three-cylinder, 1.0 liter

engine.


32.7, 35.7, 38.0, 38.7, 40.1, 42.3, 34.3, 36.2, 38.1, 39.0, 40.6, 42.9, 34.5, 37.4, 38.3, 39.4, 41.4, 43.4, 35.3, 37.6, 38.5, 39.7, 41.7, 48.9





(a) Compute the z-score corresponding to the individual who obtained 38.1 miles per gallon. Interpret this result.


The z-score corresponding to the individual is __ and indicated the date value is __ standard decoration(s) __(below/above) the __(mean/median)

(b) Determine the quartiles.

(c) Compute and interpret the interquartile range, IQR.

(d) Determine the lower and upper fences. Are there any outliers?

Answer:

A) Z = -0.24

We can interpret this as the z-score corresponding to the individual who obtained 38. 1 miles per gallon is -0.24.

B) Q1 = 36.8 and Q3 = 41

C)IQR = 4.2

This means that the distance between the first and third quartile is 4.2

D) LF = 30.5 and UF = 47.3

the only outlier is 48.9 which is more than UF

Step-by-step explanation:

We are given the data representing the miles per gallon of a random sample of cars with a three-cylinder as;

32.7, 35.7, 38.0, 38.7, 40.1, 42.3, 34.3, 36.2, 38.1, 39.0, 40.6, 42.9, 34.5, 37.4, 38.3, 39.4, 41.4, 43.4, 35.3, 37.6, 38.5, 39.7, 41.7, 48.9

Sum of data is;

ΣX = 32.7 + 35.7 + 38.0 + 38.7 + 40.1 + 42.3 + 34.3 + 36.2 + 38.1 + 39.0 + 40.6 + 42.9 + 34.5 + 37.4 + 38.3 + 39.4 + 41.4 + 43.4 + 35.3 + 37.6 + 38.5 + 39.7 + 41.7 + 48.9 = 934.7

Formula for mean is;

μ = ΣX/n

μ = 934.7/24

μ = 38.95

Formula for sample standard deviation of is;

s = √[(1/(n - 1))Σ(X - μ)²]

From online calculator, we have a value of;

s = 3.505

A) Formula for z-score is;

Z = (x - μ)/s

We are given x = 38.1

Thus;

Z = (38.1 - 38.95)/3.505

Z = -0.24

We can interpret this as the z-score corresponding to the individual who obtained 38. 1 miles per gallon is -0.24.

B) Arranging the set of given values in ascending order, we have;

32.7, 34.3, 34.5, 35.3, 35.7, 36.2, 37.4, 37.6, 38.0, 38.1, 38.3, 38.5, 38.7, 39.0, 39.4, 39.7, 40.1, 40.6, 41.4, 41.7, 42.3, 42.9, 43.4, 48.9

Number of data; n = 24

First quartile is;

Q1 = ((n + 1)/4)th = ((24 + 1)/4)th = 25/4 = 6.25th term

Thus, we will find the average of the 6th and 7th term;

Q1 = (36.2 + 37.4)/2

Q1 = 36.8

Similarly;

Q3 = ¾(n + 1)th = ¾(24 + 1)

Q3 = ¾ × 25 = 18.75th term

Thus, we will find the average of the 18th and 19th term.

Q3 = (40.6 + 41.4)/2

Q3 = 41

C) Formula for inter quartile range is;

IQR = Q3 - Q1

IQR = 41 - 36.8

IQR = 4.2

This means that the distance between the first and third quartile is 4.2

D) Formula for lower fence is;

LF = Q1 - (1.5 × IQR)

LF = 36.8 - (1.5 × 4.2)

LF = 30.5

Formula for upper fence is;

UF = Q3 + (1.5 × IQR)

UF = 41 + (1.5 × 4.2)

UF = 47.3

Now, an outlier will be any values in the given set less than LF or more than UF.

From the given set, the only outlier is 48.9 which is more than UF

PLEASE ANSWER!!!! I WILL MARK YOU AS BRAINLIEST PLEASE SCHOOL IS STRESSFUL AND I JUST WANT TO GET THIS DONE!!!

Help me out? Please! Thank you!

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