Respuesta :
Answer:
(1) The probability that the time to deliver a pizza is at least 32 minutes is 0.70.
(2a) The percentage of results more than 45 is 79.67%.
(2b) The percentage of results less than 85 is 91.77%.
(2c) The percentage of results are between 75 and 90 is 15.58%.
(2d) The percentage of results outside the healthy range 20 to 100 is 2.64%.
Step-by-step explanation:
(1)
Let Y = the time taken to deliver a pizza.
The random variable Y follows a Uniform distribution, U (20, 60).
The probability distribution function of a Uniform distribution is:
[tex]f(x)=\left \{ {{\frac{1}{b-a};\ x\in [a, b] } \atop {0};\ otherwise} \right.[/tex]
Compute the probability that the time to deliver a pizza is at least 32 minutes as follows:
[tex]P(Y\geq 32)=\int\limits^{60}_{32} {\frac{1}{b-a} } \, dx \\=\frac{1}{60-20} \int\limits^{60}_{32} {1 } \, dx\\=\frac{1}{40}\times[x]^{60}_{32}\\=\frac{1}{40}\times[60-32]\\=0.70[/tex]
Thus, the probability that the time to deliver a pizza is at least 32 minutes is 0.70.
(2)
Let X = results of a certain blood test.
It is provided that the random variable X follows a Normal distribution with parameters [tex]\mu = 60[/tex] and [tex]s = 18[/tex].
The probabilities of a Normal distribution are computed by converting the raw scores to z-scores.
The z-scores follows a Standard normal distribution, N (0, 1).
(a)
Compute the probability that the results are more than 45 as follows:
[tex]P(X>45)=P(\frac{X-\mu}{\sigma}> \frac{45-60}{18})=P(Z>-0.833)=P(Z<0.833)=0.7967[/tex]
The percentage of results more than 45 is: [tex]0.7967\times100=79.67\%[/tex]
Thus, the percentage of results more than 45 is 79.67%.
(b)
Compute the probability that the results are less than 85 as follows:
[tex]P(X<85)=P(\frac{X-\mu}{\sigma}< \frac{85-60}{18})=P(Z<1.389)=0.9177[/tex]
The percentage of results less than 85 is: [tex]0.9177\times100=91.77\%[/tex]
Thus, the percentage of results less than 85 is 91.77%.
(c)
Compute the probability that the results are between 75 and 90 as follows:
[tex]P(75<X<90)=P(\frac{75-60}{18}<\frac{X-\mu}{\sigma}< \frac{90-60}{18})\\=P(0.833<Z<1.67)\\=P(Z<1.67)-P(Z<0.833)\\=0.9525-0.7967\\=0.1558[/tex]
The percentage of results are between 75 and 90 is: [tex]0.1558\times100=15.58\%[/tex]
Thus, the percentage of results are between 75 and 90 is 15.58%.
(d)
Compute the probability that the results are between 20 and 100 as follows:
[tex]P(20<X<100)=P(\frac{20-60}{18}<\frac{X-\mu}{\sigma}< \frac{100-60}{18})\\=P(-2.22<Z<2.22)\\=P(Z<2.22)-P(Z<-2.22)\\=0.9868-0.0132\\=0.9736[/tex]
Then the probability that the results outside the range 20 to 100 is: [tex]1-0.9736=0.0264[/tex].
The percentage of results outside the range 20 to 100 is: [tex]0.0264\times100=2.64\%[/tex]
Thus, the percentage of results outside the healthy range 20 to 100 is 2.64%.
