Question

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.

Answer 1

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 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, $X\sim N(129.71,\ 2.28^{2})$.

(a)

Compute the probability that a lap is completes in less than 130 seconds as follows:

$P(X$

                   $=P(Z$

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:

$z=\frac{x-\mu}{\sigma}\\-1.88=\frac{x-129.71}{2.28}\\x=129.71-(1.88\times 2.28)\\x=125.4236\\x\approx 125.42$

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,

$P(x_{1}$

The value of z for the above probability is:

z = 1.28

Compute the values of x₁ and x₂ as follows:

$-z=\frac{x_{1}-\mu}{\sigma}\\-1.28=\frac{x_{1}-129.71}{2.28}\\x_{1}=129.71-(1.28\times 2.28)\\x=126.7916\\x\approx 126.80$               $z=\frac{x_{2}-\mu}{\sigma}\\1.28=\frac{x_{2}-129.71}{2.28}\\x_{2}=129.71+(1.28\times 2.28)\\x=132.6284\\x\approx 132.63$

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