The amount of insurance​ (in thousands of​ dollars) sold in a day by a particular agent is uniformly distributed over the interval ​[5​, 70​]. A. What amount of insurance does the agent sell on an average​ day? B. Find the probability that the agent sells more than ​$40​,000 of insurance on a particular day.

A. The agent sells ​$ nothing of insurance on an average day.
B. The probability that the agent sells more than ​$40​,000 of insurance on a particular day is nothing. ​(Round to two decimal places as​ needed.)

Respuesta :

Answer:

a) The agent sells ​37,500$ of insurance on an average day.

b) There is a 53.85% probability that the agent sells more than ​$40​,000 of insurance on a particular day.

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability that we find a value X lower than x is given by the following formula.

[tex]P(X \leq x) = \frac{x - a}{b-a}[/tex]

The mean of this distribution is given by the following formula:

[tex]M = \frac{a+b}{2}[/tex]

For this problem, we have that:

[tex]a = 5000, b = 70000[/tex]

A. What amount of insurance does the agent sell on an average​ day?

[tex]M = \frac{5000 + 70000}{2} = 37500[/tex]

The agent sells ​37,500$ of insurance on an average day.

B. Find the probability that the agent sells more than ​$40​,000 of insurance on a particular day.

[tex]P(X \leq 40000) + P(X > 40000) = 1[/tex]

[tex]P(X > 40000) = 1 - P(X \leq 40000) = 1 - \frac{40000 - 5000}{70000 - 5000} = 0.5385[/tex]

There is a 53.85% probability that the agent sells more than ​$40​,000 of insurance on a particular day.