Application of Dispersion

2011-05-25 3:36 am

A company produces boxes of drawing pins which are claimed to contain at least 40pins per box. The company knows that the number of pins in each box follow a normal distribution with a mean of 44pins per box and a Standard Deviation of 2pins.

1) What percentage of the boxes of drawing pins produced will contain less than 40 pins?

2) If the company has to supply 3900boxes of drawing pins to the retailers per day, and the number of drawing pins in each box must exceed 40, how many boxes of drawing pins should be produced in order to fulfill the need?

3) If the operation of the machines are improved and the Standard Deviation is decrease to 1pin, find the mean number of pins that a box should contain so that the percentage of the boxes which contain less than 40pins remain the same as before.

回答 (1)

用戶10012

2011-05-26 4:37 pm

✔ 最佳答案

(1) Let X = No. of pins in a box. So X ~ N(44, 4).
P(Less than 40 pins) = P(X < 40) = P[(X - 44)/2 < (40 - 44)/2]
= P(Z < - 2) = 0.5 - A(2) = 0.5 - 0.4772 = 0.0228 = 2.28%.
(Note : A(2) = 0.4772 is obtained from the distribution table.)
(2) P(More than 40 pins) = 1 - 0.0228 = 0.9772
So pins to be produced = 3900/0.9772 = 3991 boxes.
(3) Let new mean no. be u, new s.d. = 1.
Let Y = No. of pins in a box. So Y ~ N(u, 1)
P(Less than 40 pins) = P(Y < 40) = P[(Y - u)/1 < (40 - u)/1]
= P[Z < (40 - u)] which is equal to 0.0228
That is 0.0228 + A[-(40 - u)] = 0.5
A(u - 40) = 0.5 - 0.0228 = 0.4772 = A(2)
u - 40 = 2
u = 42.
So new mean is 42.

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