Answer:
a) P(x<5)=0.
b) E(X)=15.
c) P(8<x<13)=0.3.
d) P=0.216.
e) P=1.
Step-by-step explanation:
We have the function:
[tex]f(x)=\left \{ {{\frac{1}{10},\, \, \, 10\leq x\leq 20 } \atop {0, \, \, \, \, \, \, otherwise }} \right.[/tex]
a) We calculate the probability that you need less than 5 minutes to get up:
[tex]P(x<5)=\int_0^5 f(x)\, dx\\\\P(x<5)=\int_0^5 0\, dx\\\\P(x<5)=0[/tex]
Therefore, the probability is P(x<5)=0.
b) It takes us between 10 and 20 minutes to get up. The expected value is to get up in 15 minutes.
E(X)=15.
c) We calculate the probability that you will need between 8 and 13 minutes:
[tex]P(8\leq x\leq 13)=P(10\leqx\leq 13)\\\\P(8\leq x\leq 13)=\int_{10}^{13} f(x)\, dx\\\\P(8\leq x\leq 13)=\int_{10}^{13} \frac{1}{10} \, dx\\\\P(8\leq x\leq 13)=\frac{1}{10} \cdot [x]_{10}^{13}\\\\P(8\leq x\leq 13)=\frac{1}{10} \cdot (13-10)\\\\P(8\leq x\leq 13)=\frac{3}{10}\\\\P(8\leq x\leq 13)=0.3[/tex]
Therefore, the probability is P(8<x<13)=0.3.
d) We calculate the probability that you will be late to each of the 9:30am classes next week:
[tex]P(x>14)=\int_{14}^{20} f(x)\, dx\\\\P(x>14)=\int_{14}^{20} \frac{1}{10} \, dx\\\\P(x>14)=\frac{1}{10} [x]_{14}^{20}\\\\P(x>14)=\frac{6}{10}\\\\P(x>14)=0.6[/tex]
You have 9:30am classes three times a week. So, we get:
[tex]P=0.6^3=0.216[/tex]
Therefore, the probability is P=0.216.
e) We calculate the probability that you are late to at least one 9am class next week:
[tex]P(x>9.5)=\int_{10}^{20} f(x)\, dx\\\\P(x>9.5)=\int_{10}^{20} \frac{1}{10} \, dx\\\\P(x>9.5)=\frac{1}{10} [x]_{10}^{20}\\\\P(x>9.5)=1[/tex]
Therefore, the probability is P=1.
