poisson distribution?

[匿名]

2015-12-02 2:55 pm

Wooden rectangular blocks are produced by a machine. The length and width of the blocks are normally distributed and their means and standard deviations are shown below:
Mean Standard deviation
Length of block 30cm 0.5cm
Width of block 10cm 0.2cm
The two distributions are independent of each other.
(a) Calculate the probability for each of the following cases:
i. The length of a randomly selected block lies within 29.35 cm and 30.75 cm.
ii. The width of a randomly selected block lies within 9.48 cm and 10.42 cm.
(b) A block will only be accepted if both its length and width lie within the limits in part (a).
i. Calculate the probability that out of 10 blocks, at least 8 will be accepted.
ii. Using normal approximation to binomial probability, calculate the probability that out of 1000 blocks, at least 800 will be accepted

回答 (1)

Lopez

2015-12-02 5:47 pm

✔ 最佳答案

(a) ( i )
P( 29.35 < L < 30.75 )
= P( (29.35-30)/0.5 < Z < (30.75-30)/0.5 ) , where Z ~ N(0,1)
= P( - 1.3 < Z < 1.5 )
= P( Z < 1.5 ) - P( Z < - 1.3 )
= 0.9332 - 0.0968
= 0.8364 ..... Ans

( ii )
P( 9.48 < W < 10.42 )
= P( (9.48-10)/0.2 < Z < (10.42-10)/0.2 )
= P( - 2.6 < Z < 2.1 )
= P( Z < 2.1 ) - P( Z < - 2.6 )
= 0.9821 - 0.0047
= 0.9774 ..... Ans

(b) ( i )
Suppose that out of 10 blocks, X blocks will be accepted.
p = 0.8364*0.9774 = 0.81749736
q = 1 - p = 0.18250264
P(x) ≡ P( X = x ) = C(10,x) * p^x * q^(10-x)

P( X ≧ 8 )
= P(8) + P(9) + P(10)
= C(10,8)*p^8*q^2 + C(10,9)*p^9*q^1 + C(10,10)*p^10
= 45*p^8*q^2 + 10*p^9*q^1 + p^10
≒ 0.730 ..... Ans

( ii )
Suppose that out of 1000 blocks, X blocks will be accepted.
p = 0.81749736
q = 1 - p = 0.18250264
μ = np = 1000*0.81749736 = 817.49736
σ = √( npq ) = √( 817.49736*0.18250264 ) ≒ 12.21456

P( X ≧ 800 )
= P( Z > (800-817.49736)/12.21456 )
≒ P( Z > - 1.43 )
= 1 - P( Z < - 1.43 )
= 1 - 0.0764
= 0.9236 ..... Ans

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