About 99.7% of vehicles whose speeds are between 59 miles per hour and 77 miles per hour.
Empirical rule states that for a normal distribution, 68% lie within one standard deviations, 95% lie within two standard deviations, and 99.7% lie within three standard deviations of the mean.
Given that mean (μ) = 68 miles per hour, standard deviation (σ) = 3 miles per hour.
68% lie within one standard deviation = (μ ± σ) = (68 ± 3) = (65, 71).
Hence 68% of the vehicle speed is between 65 miles per hour and 71 miles per hour.
95% lie within two standard deviation = (μ ± 2σ) = (68 ± 2*3) = (62, 74).
Hence 95% of the vehicle speed is between 62 miles per hour and 74 miles per hour.
99.7% lie within three standard deviation = (μ ± 3σ) = (68 ± 3*3) = (59, 77).
Hence 99.7% of the vehicle speed is between 59 miles per hour and 77 miles per hour.
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The question is an illustration of probabilities from z score.
The percentage of vehicles whose speeds are between 59 miles per hour and 77 miles per hour is 99.73%
The given parameters are:
[tex]\mu = 68[/tex] -- the mean
[tex]\sigma = 3[/tex] --- the standard deviation
The percent of vehicles between 59 and 77 miles per hour is represented as:
[tex]x_1 = 59[/tex]
[tex]x_2 = 77[/tex]
So, the probability is represented as:
[tex]P(59 < x < 77) = P(z_1 < z < z_2)[/tex]
Calculate the z values using:
[tex]z = \frac{x - \mu}{\sigma}[/tex]
When [tex]x_1 = 59[/tex]
[tex]z_1 = \frac{59 - 68}{3} = -3[/tex]
When [tex]x_2 = 77[/tex]
[tex]z_2 = \frac{77 - 68}{3} = 3[/tex]
So, we have:
[tex]P(59 < x < 77) = P(z_1 < z < z_2)[/tex]
[tex]P(59 < x < 77) = P(-3 < z < 3)[/tex]
Using empirical rule:
[tex]P(-3<z<3) = 0.9973[/tex] --- from z table of probabilities
So, we have:
[tex]P(59<x<77) = 0.9973[/tex]
Represent as percentage
[tex]P(59<x<77) = 0.9973 \times 100\%[/tex]
[tex]P(59<x<77) = 99.73\%[/tex]
Hence, the percentage of vehicles whose speeds are between 59 miles per hour and 77 miles per hour is 99.73%
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https://brainly.com/question/16464000
