Probability Distributions!!

[匿名]

2010-05-19 12:04 am

If you toss a fair coin 100 times, what is the probability that you will get
a) at least 40 heads?
b) between 40 to 70 heads?

The Coca-cola company is coming out with new 3-liter packaging. The actual filling in each bottle is normally distributed, with a mean of 3 liters and standard deviation of 0.08 liter.
a) What percentage of the bottles will contain between 2.90 liters and 3.10 liters?
b) 98% of the bottles contain at least how much mineral water?
c) 98% of the bottles contain an amount that is between which two values around the mean?

Please help me! thx a lot!!!

回答 (1)

Prof. Physics

2010-05-19 3:17 am

✔ 最佳答案

1. It is a binomial distribution

X ~ B(100, 1/2)

Under normal approximation, X ~ N(100(1/2), 100(1/2)(1/2)) ~ N(50, 5^2)

a. P(X >= 40)

= P[z >= (40 - 0.5 - 50)/5] (as it is a discrete distribution)

= P(z >= -2.1)

= 0.5 + 0.4821

= 0.9821

b. P[40 <= X <= 70]

= P[(40 - 0.5 - 50)/5 <= z <= (70 + 0.5 - 50)/5]

= P(-2.1 <= 4.1)

= 0.5 + 0.4821

= 0.9821 (Essentially it is the same as the answer in (a))

2.a. Let X be the volume of the package

X ~ N(3, 0.08^2)

a. P(2.90 <= X <= 3.10)

= P[(2.90 - 3)/0.08 <= z <= (3.10 - 3)/0.08]

= P(-1.25 <= z <= 1.25)

= 2 X 0.3944

= 0.7888

b. P(X > c) = 0.98

P[z > (c - 3)/0.08] = 0.98

(c - 3)/0.08 = -2.054

c = 2.8357 litres

c. Test statistics, z = 2.32666667

So, the required bound

= (3 - 2.3266667 X 0.08, 3 + 2.3266667 X 0.08)

= (2.8139, 3.1861)

參考: Prof. Physics

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