又是一題概率的題目 = = (大學)

[匿名]

2009-10-19 7:35 pm

The width of a casing for a door is normally distributed with a mean of 24 inches and a standard deviation of 1/8 inch. The width of a door is normally distributed with a mean of 23 and 7/8 inches and a standard deviation of 1/16 inch. Assume independence.

a.
Determine the mean and standard deviation of the difference between the width of the casing and the width of the door.

b.
What is the probability that the width of the casing minus the width of the door exceeds 1/4 inch?

c.
What is the probability that the door does not fit in the casing?

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回答 (1)

Prof. Physics

2009-10-20 5:36 am

✔ 最佳答案

a. Let X and Y be the random variables of the width of door case and the width of the door respectively.

X ~ N(24 , (1/8)^2)

Y ~ N(191/8 , (1/16)^2)

Difference between the width of the casing and the width of the door = X - Y

Assume the difference also follows normal distribution.

X - Y ~ N(24 - 191/8 , (1/8)^2 + (1/16)^2)

X - Y ~ N(1/8 , 5/256)

X - Y ~ N(1/8 , (sqrt5 / 16)^2)

So, the mean of the difference is 1/8 inches, and the standard deviation is (sqrt5) / 16

b. Let M be the random variable of the width of the casing minus the width of the door.

M = X - Y ~ N(1/8 , (sqrt5 / 16)^2)

P(M > 1/4)

= P[z > (1/4 - 1/8)/(sqrt5 / 16)]

= P(z > 0.894427191)

= 0.5 - 0.3145

= 0.1855

c. For the door does not fit in the casing, the width of the door is larger than the width of the casing.

X - Y < 0

P(M < 0)

= P[z < (0 - 1/8)/(sqrt5 / 16)]

= P(z < -0.894427191)

= 0.5 - 0.3145

= 0.1855

參考: Physics king

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