Answer: 1
Step-by-step explanation:
If a random variable x is uniformly distributed in [a,b] the
Mean = [tex]\dfrac{a+b}{2}[/tex]
Standard deviation : [tex]\sqrt{\dfrac{(b-a)^2}{12}}[/tex]
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 = [tex]\dfrac{65+40}{2}=52.5[/tex] minutes
Standard deviation : [tex]\sqrt{\dfrac{(65-40)^2}{12}}\approx7.22[/tex]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.
The empirical rule states that 95% of the distribution is within 2 standard deviations of the mean,
Z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:
z = (x - μ)/σ
where x is the raw score, μ is the mean and σ is the standard deviation.
The empirical rule states that 95% of the distribution is within 2 standard deviations of the mean.
Find out more on Z score at: https://brainly.com/question/25638875
