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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?

Respuesta :

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

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