On the way home, Mark rode the first six miles of the trip in 30
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Answer:
(a) The percent of her laps that are completed in less than 130 seconds is 55%.
(b) The fastest 3% of her laps are under 125.42 seconds.
(c) The middle 80% of her laps are from <u>126.80</u> seconds to <u>132.63</u> seconds.
Step-by-step explanation:
The random variable <em>X</em> is defined as the number of seconds for a randomly selected lap.
The random variable <em>X </em>is normally distributed with mean, <em>μ</em> = 129.71 seconds and standard deviation, <em>σ</em> = 2.28 seconds.
Thus, .
(a)
Compute the probability that a lap is completes in less than 130 seconds as follows:
The percentage is, 0.55 × 100 = 55%.
Thus, the percent of her laps that are completed in less than 130 seconds is 55%.
(b)
Let <em>x</em> represents the 3rd percentile.
That is, P (X < x) = 0.03.
⇒ P (Z < z) = 0.03
The value of <em>z</em> for the above probability is:
<em>z</em> = -1.88
Compute the value of <em>x</em> as follows:
Thus, the fastest 3% of her laps are under 125.42 seconds.
(c)
Let <em>x</em>₁ and <em>x</em>₂ be the values between which the middle 80% of the distribution lie.
That is,
The value of <em>z</em> for the above probability is:
<em>z</em> = 1.28
Compute the values of <em>x</em>₁ and <em>x</em>₂ as follows:
Thus, the middle 80% of her laps are from <u>126.80</u> seconds to <u>132.63</u> seconds.