A technician services mailing machines at companies in the Phoenix area. Depending on the type of malfunction, the service call can take 1.2, 2.3, 3.3, or 4.1 hours. The different types of malfunctions occur at the same frequency. If required, round your answers to two decimal places. Develop a probability distribution for the duration of a service call.

Duration of Call x f(x)
1.2 ____
2.3 ____
3.3 ____
4.1 ____
Consider the required conditions for a discrete probability function, shown below.
f(x)>/=0 (5.1)
summation f(x)=1 (5.2)
Does this probability distribution satisfy equation (5.1)? _________________
Does this probability distribution satisfy equation (5.2)? _________________
What is the probability a service call will take 3.3 hours? ____
A service call has just come in, but the type of malfunction is unknown. It is 3:00 P.M. and service technicians usually get off at 5:00 P.M. What is the probability the service technician will have to work overtime to fix the machine today? ____

Respuesta :

Answer:

(a) The probability distribution table is provided in the attachment.

(b) The probability distribution of X is discrete.

(c) The probability a service call will take 3 hours is 0.30.

(d) The probability that the service technician will have to work overtime is 0.70.

Step-by-step explanation:

Let X = number of hours a service calls take.

(a)

The probability of an event E is:

[tex]P(E)=\frac{n(E)}{N}[/tex]

Here,

n (E) = number of favorable outcomes

N = total number of outcomes.

The probability distribution table is provided in the attachment.

(b)

For a discrete probability distribution:

  • [tex]f(x) \geq 0\\[/tex]
  • [tex]\sum f(x)=1[/tex]

Check that the probability distribution of X is discrete as follows:

The value of f (x) for all values of x is greater than 0.

The sum of all f (x) is 1.

Thus, the probability distribution of X is discrete.

(c)

The probability of X = 3 is:

[tex]P(X=3) = f(3) = 0.30[/tex]

Thus, the probability a service call will take 3 hours is 0.30.

(d)

The service technician has 2 hours (3:00 PM to 5:00 PM) to fix the machine.

Compute the probability that the service technician will have to work overtime as follows:

P (X > 2) = P (X = 3) + P (X = 4)

              [tex]=0.30+0.40\\=0.70[/tex]

Thus, the probability that the service technician will have to work overtime is 0.70.