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aleksandr82 [10.1K]
3 years ago
8

You always need some time to get up after the alarm has rung. You get up from 10 to 20 minutes later, with any time in that inte

rval being equally likely. a. What is the probability that you need less than 5 minutes to get up?b. What is the expected amount of minutes that you will need to get up? Interpret this in the context of the problem in a sentence or two. c. What is the probability that you will need between 8 and 13 minutes? You have 9:30am classes three times a week and 9am classes two times a week, with each class being independent from the other. d. You will be late to your 9:30am class if you need more than 14 minutes to get up. What is the probability that you will be late to each of the 9:30am classes next week? e. You will be late to your 9am class if you need more than 9 and a half minutes to get up. What is the probability that you are late to at least one 9am class next week?
Mathematics
1 answer:
Mama L [17]3 years ago
3 0

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:

f(x)=\left \{ {{\frac{1}{10},\, \, \, 10\leq x\leq 20 } \atop {0, \, \, \, \, \, \,  otherwise }} \right.

a)  We calculate  the probability that you need less than 5 minutes to get up:

P(x

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:

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

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:

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

You have 9:30am classes three times a week.  So, we get:

P=0.6^3=0.216

Therefore, the probability is P=0.216.

e)  We calculate the probability that you are late to at least one 9am class next week:

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

Therefore, the probability is P=1.

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