Question

Suppose that the times required for a cable company to fix cable problems in the homes of its customers are uniformly distributed between 40 minutes and 65 minutes. What is the probability that a randomly selected cable repair visit falls within 2 standard deviations of the mean?

Answer 1

Answer: 1

Step-by-step explanation:

If a random variable x is uniformly distributed in [a,b] the

Mean = $\dfrac{a+b}{2}$

Standard deviation : $\sqrt{\dfrac{(b-a)^2}{12}}$

Let x =  Times required for a cable company to fix cable problems

As per given.

x is  uniformly distributed between 40 minutes and 65 minutes.

Then , mean = $\dfrac{65+40}{2}=52.5$ minutes

Standard deviation : $\sqrt{\dfrac{(65-40)^2}{12}}\approx7.22$minutes

Consider , P (mean- 2(Standard deviation) < X < mean+2(Standard deviation) )

= P(52.5-2(7.22)< X <  52.5+2(7.22))

=P(38.06 <X < 66.94 ).

But x lies between 40 minutes and 65 minutes.

Also,   [40 minutes,  65 minutes]⊂ [38.06 minutes , 66.94 minutes]

Therefore ,P(38.06 <X < 66.94 ) =1

∴ The probability that a randomly selected cable repair visit falls within 2 standard deviations of the mean is 1.