Answer:
The z- score corresponding to the individual who obtained 42.3 miles per gallon is 0.9474 which indicates that it is 0.9474 standard deviations below the mean.
Step-by-step explanation:
From the data set attached below;
Let consider that the random variable X as the miles per gallon driven by cars with a three cylinders, 1.0 liter engine.
The mean [tex]\mu[/tex] of the data set can then be calculated as :
[tex]\mu = \dfrac{1}{n} \sum\limits^n_{ i=1} x_i[/tex]
= [tex]\dfrac{31.5+34.2+34.7+...+42.5+43.4+49.3}{24}[/tex]
= 38.88
The standard deviation is calculated by the formula:
[tex]\sigma = \sqrt{\dfrac{1}{n-1} \sum\limits^n_{ i=1} (x_i- \mu)^2}[/tex]
= [tex]\sqrt{\dfrac{(31.5-38.88)^2+...+(49.3-38.88)^2}{24-1}}[/tex]
= 3.61
The z score corresponding to the value x = 42.3 can be calculated as:
[tex]z = \dfrac{x-\mu}{\sigma}[/tex]
[tex]z = \dfrac{42.3-38.88}{3.61}[/tex]
z = 0.9474
The z- score corresponding to the individual who obtained 42.3 miles per gallon is 0.9474 which indicates that it is 0.9474 standard deviations below the mean.
