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請這題數點計嫁?

用戶104333

2011-10-31 12:36 am

A machine discharges soda water once for each cup of soda water purchased. The amount of soda water in each discharge is independently normally distributed with mean 300 mL and standard deviation 5 mL. (a) What is the probability that the amount of a cup of soda water is (i) more than 308 mL? (ii) between 295 and 315 mL? (b) The machine is adjusted so that only 95% soda water contain more than k mL, find the value of k. (c) Today, 25 cups of soda water is discharged by the machine, what is the probability that 2 or more contain more than 308 mL.

回答 (1)

用戶10012

2011-10-31 5:12 pm

✔ 最佳答案

Let X = Amount of water discharged.
(a) P(Amount more than 308) = P(X > 308) = P[(X - 300)/5 > (308 - 300)/5]
= P(Z > 1.6) = 0.5 - A(1.6) = 0.5 - 0.4452 = 0.0548.
(b) P(Amount between 295 and 315) = P(295 < X < 315)
= P[(295 - 300)/5 < (X - 300)/5 < (315 - 300)/5]
= P(- 1 < Z < 3) = A(1) + A(3) = 0.3413 + 0.4987 = 0.84.
(c) P(Amount more than k) = P(X > k)
= P[(X - 300)/5 > (k - 300)/5] = P[Z > (k - 300)/5] = A[(k - 300)/5] which is 95%.
95% = 0.95 = 0.5 + 0.45 = 0.5 + A( - 1.645)
so A(-1.645) = A[(k - 300)/5]
- 1.645 = (k - 300)/5
k = 300 - 8.225 = 291.775 mL.
(c) This is a Binomial Distribution with n = 25 and p = 0.0548
P(2 or more cups contain more than 308) = 1 - P(No cup more than 308) - P(1 cup more than 308) = 1 - 25C0(1 - 0.0548)^25 - 25C1(0.0548)(1 - 0.0548)^24
= 1 - 0.2444 - 0.3542 = 0.40136

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