Math Probability statistics

JetGot

2013-10-06 5:30 am

1) A research scientist reports that mice will live an average of 40 months when their diets are sharply restricted and then enriched with vitamins and proteins. Assuming that the life times of such miceare normally distributed with a standard deviation of 6.3 months, find the probability that a given mouse will live

a)more than 32 months;
(b)less than 28 months;
(c)between 37 and 49 months

2) A soft-drink machine is regulated so that it discharges an average of 200 milliliters per cup. If the amount of drink is normally distributed with a standard deviation equal to 15 milliliters:

a)what fraction of the cups will contain more than 224 milliliters?
(b)what is the probability that a cup contains between 191 and 209 milliliters?
(c)how many cups will probably over flow if 230 milliliter cups are used for the next 1000 drinks?
(d)below what value do we get the smallest 25% of the drinks

The probability that a patient recovers from a delicate heart operation is 0.9.Of the next 100 patients having this operation, what is the probability that
(a)between 84 and 95 inclusive survive?
(b)fewer than 86 survive

回答 (1)

myisland8132

2013-10-06 9:01 pm

✔ 最佳答案

1(a) P(X > 32)

= P(Z > (32 - 40)/6.3)

= P(Z > -1.2698)

= 0.8979

(b) P(X < 28)

= P(Z < (28 - 40)/6.3)

= P(Z > -1.9048)

= 0.0284

(c) P(37 <= X <= 49)

= P((37 - 40)/6.3 <= Z <= (49 - 40)/6.3)

= P(-0.4762 <= Z <= 1.4286)

= 0.6065

2(a) P(X > 224)

= P(Z > (224 - 200)/15)

= P(Z > 1.6)

= 0.0548

(b) P(191 <= X <= 209)

= P((191 - 200)/15 < Z < (209 - 200)/15)

= P(-0.6 < Z < 0.6)

= 0.4515

(c) P(X > 230)

= P(Z > (230 - 200)/15)

= P(Z > 2)

= 0.0228

So, 22 cups will be over flow if 230 milliliter cups are used for the next 1000 drinks

(d) 200 - 1.96 * 15 = 170.6

3(a) mean = 90 and standard deviation = sqrt(100 * 0.9 * 0.1) = 3

P(84 <= X <= 95)

= P(-5.3333 <= Z <= -3.4444)

= 0.0003

(b) P(X < 86)

= P(Z < -4.6667)

~ 0

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