Population mean?

2017-08-12 12:27 pm

The life span of a rabbit has a normal distribution with mean of 9.2 years and standard deviation of 1.8 years. A random sample of 36 rabbits is selected, find the probability that the mean lifespan is
(a) within three quarters of a standard deviation of the population mean
(b) within 0.5 year of population mean
(c) at least 0.5 year lower than population mean
(d) between 8.1 and 9.3 years
How to solve this? Can show me the steps?

回答 (1)

Lopez

2017-08-12 9:05 pm

✔ 最佳答案

(a)
μ of xbar = μ = 9.2
σ of xbar = σ / √n = 1.8 / √36 = 1.8 / 6 = 0.3

P( 9.2 - 1.8*3/4 < xbar < 9.2 + 1.8*3/4 )
= P( 9.2 - 1.35 < xbar < 9.2 + 1.35 )
= P( - 1.35 < xbar - 9.2 < 1.35 )
= P( - 1.35/0.3 < ( xbar - 9.2 )/0.3 < 1.35/0.3 )
= P( - 4.5 < Z < 4.5 ) , where Z ~ N(0,1)
= 1 - 2*P( Z < - 4.5 )
≒ 1 - 2*3.4*10^(-6)
≒ 0.999993 ..... Ans

(b)
P( - 0.5 < xbar - 9.2 < 0.5 )
= P( - 0.5/0.3 < ( xbar - 9.2 )/0.3 < 0.5/0.3 )
≒ P( - 1.67 < Z < 1.67 )
= 1 - 2*( Z < - 1.67 )
= 1 - 2*0.0475
= 0.905 ..... Ans

(c)
P( xbar < 9.2 - 0.5 )
= P( xbar - 9.2 < - 0.5 )
= P( ( xbar - 9.2 )/0.3 < - 0.5/0.3 )
≒ P( Z < - 1.67 )
= 0.0475 ..... Ans

(d)
P( 8.1 < xbar < 9.3 )
= P( 8.1 - 9.2 < xbar - 9.2 < 9.3 - 9.2 )
= P( - 1.1 < xbar - 9.2 < 0.1 )
= P( - 1.1/0.3 < ( xbar - 9.2 )/0.3 < 0.1/0.3 )
≒ P( - 3.67 < Z < 0.33 )
= P( Z < 0.33 ) - P( Z < - 3.67 )
≒ 0.6293 - 1.2*10^(-4)
≒ 0.629 ..... Ans

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