Respuesta :
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, [tex]X\sim N(129.71,\ 2.28^{2})[/tex].
(a)
Compute the probability that a lap is completes in less than 130 seconds as follows:
[tex]P(X<130)=P(\frac{X-\mu}{\sigma}<\frac{130-129.71}{2.28})[/tex]
[tex]=P(Z<0.13)\\=0.55172\\\approx0.55[/tex]
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:
[tex]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[/tex]
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,
[tex]P(x_{1}<X<x_{2})=0.80\\P(-z<Z<z)=0.80\\P(Z<z)-P(Z<-z)=0.80\\P(Z<z)-[1-P(Z<z)]=0.80\\2P(Z<z)=1.80\\P(Z<z)=0.90[/tex]
The value of z for the above probability is:
z = 1.28
Compute the values of x₁ and x₂ as follows:
[tex]-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[/tex] [tex]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[/tex]
Thus, the middle 80% of her laps are from 126.80 seconds to 132.63 seconds.
