Terri Vogel, an amateur motorcycle racer, averages 129.71 seconds per 2.5 mile lap (in a 7 lap race) with a standard deviation of 2.28 seconds . The distribution of her race times is normally distributed. We are interested in one of her randomly selected laps. (Source: log book of Terri Vogel) Let X be the number of seconds for a randomly selected lap. Round all answers to two decimal places.

(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 126.80 seconds to 132.63 seconds.

Step-by-step explanation:

The random variable X is defined as the number of seconds for a randomly selected lap.

The random variable X is normally distributed with mean, μ = 129.71 seconds and standard deviation, σ = 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 x represents the 3rd percentile.

That is, P (X < x) = 0.03.

⇒ P (Z < z) = 0.03

The value of z for the above probability is:

z = -1.88

Compute the value of x as follows:

Thus, the fastest 3% of her laps are under 125.42 seconds.

(c)

Let x₁ and x₂ be the values between which the middle 80% of the distribution lie.

That is,

The value of z for the above probability is:

z = 1.28

Compute the values of x₁ and x₂ as follows:

Thus, the middle 80% of her laps are from 126.80 seconds to 132.63 seconds.

use this formula:

p(a\b) = p(a and b)/p(b).

take it from here.

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step-by-step explanation: