Answer:
a
[tex]P(X > 7) = 0.5[/tex]
b
[tex]P(X < 3) = -0.5[/tex]
c
[tex]x = 6.2 \ hours[/tex]
d
[tex]x_1 = 8 \ hours[/tex]
e
Mean [tex]\mu = 7[/tex]
Standard deviation [tex]s \approx 1.1547[/tex]
Step-by-step explanation:
From the question we are told that
The amount of time a repairman needs to fix a furnace is uniformly distributed from 5 and 9 hours
Generally the probability that it takes the repairman more than 7 hours to fix a furnace is mathematically represented as
[tex]P(X > 7) = \frac{9- 7}{9-5}[/tex]
=>[tex]P(X > 7) = 0.5[/tex]
Generally the probability that a randomly selected furnace repair requires less than 3 hours. is mathematically represented as
[tex]P(X < 3) = \frac{3 -5}{9-5}[/tex]
=>[tex]P(X < 3) = -0.5[/tex]
Generally the 30th percentile of furnace repair times is mathematically represented as
[tex]30^{th} \ percentile = \frac{x - 5}{9-5}[/tex]
=> [tex]0.3 = \frac{x - 5}{9-5}[/tex]
=> [tex]x = 6.2 \ hours[/tex]
Generally the time taken for the longest 25% of furnace repair is mathematically evaluated as
[tex]75^{th} \ percentile = \frac{x_1 - 5}{9-5}[/tex]
Here making use of the [tex]75^{th}[/tex] because we are considering duration that fall with the top [tex]25^{th}[/tex]
=> [tex]0.75 = \frac{x_1 - 5}{9-5}[/tex]
=> [tex]x_1 = 8 \ hours[/tex]
Generally the mean is mathematically represented as
[tex]\mu = \frac{5 + 9 }{2}[/tex]
=> [tex]\mu = 7[/tex]
Generally the standard deviation is mathematically represented as
[tex]s = \sqrt{ \frac{(9- 5)^2}{12} }[/tex]
=> [tex]s \approx 1.1547[/tex]