Question

These are all part of the same question if you cant answer them all dont asnwer at all. I will thumbsdown. thanks
Let y denote the number of broken eggs in a randomly selected carton of one dozen eggs. Suppose that the probability distribution of v is as follows У01234 p(y) 0.64 0.20 0.11 0.04? (a) Only y values of 0, 1, 2, 3, and 4 have positive probabilities. What is p(4)? (Hint: Consider the properties of a discrete probability distribution.) (b) How would you interpret p(1)0.20? O The proportion of eggs that will be broken in each carton from this population is 0.20 O In the long run, the proportion of cartons that have exactly one broken egg will equal 0.20 O The probability of one randomly chosen carton having broken eggs in it is 0.20 O If you check a large number of cartons, the proportion that will have at most one broken egg will equal 0.20 (c) Calculate P(y 2), the probability that the carton contains at most two broken eggs Interpret this probability O If you check a large number of cartons, the proportion that will have at most two broken eggs will equal 0.95 O In the long run, the proportion of cartons that have exactly two broken eggs will equal 0.95 O The probability of two randomly chosen cartons having broken eggs in them is 0.95 O The proportion of eggs that will be broken in any two cartons from this population is 0.95 (d) Calculate P(y < 2), the probability that the carton contains fewer than two broken eggs Why is this smaller than the probability in Part (c) O This probability is not less than the probability in Part (c) because the two probabilities are the same for this distribution O This probability is less than the probability in Part (c) because in probability distributions, P(y s k) is always greater than P(y < k) O This probability is less than the probability in Part (c) because the proportion of eggs with any exact number of broken eggs is negligible O This probability is less than the probability in Part (c) because the event y 2 is now not included (e) What is the probability that the carton contains exactly 10 unbroken eggs? (Hint: What is the corresponding value of

Answer 1

#1)

a) Total of probabilities P(y) of all y's must be equal to 1

0.64+0.20+0.11+0.04 + P(4) = 1

0.99 +P(4) = 1

Therefore P(4) = 1 - 0.99

P(4) = 0.01

b) P(1) = 0.2

In the long run , the proportion of cartons that have exactly one broken egg will equal 0.20

c) P( y <= 2 ) = P(0) +P(1)+P(2) = 0.64+ 0.20 + 0.11

P( y <= 2 ) = 0.95

Interpretation :

If you check a large number of cartons , the proportion that will have at most two broken eggs will equal to 0.95

d) P( y < 2 ) = P(0) +(1) = 0.64+0.20 = 0.84

This probability is less than the probability in part c) because the event y=2 is now not included.

e) The carton contain total 12 eggs , so P( exactly 10 unbroken eggs) = P( 2 broken eggs )

P(y = 2 ) = 0.11

#2)

1000 P(x) 400 P(x) 0.4 0.3 0.25 0.05 10 250 Total 1000

You would expect roughly 400 of the graduates to donate nothing, roughly 300 to donate $10, roughly 250 to donate $25,and roughly 50 to donate $50. the frequencies would be close to but not exactly , these values.The four frequencies would add to 1000

b) Most common value of x in this population is $0 ( because its proportion is higher )

c)P( x >= 25 ) = P( 25 ) + P( 50 ) = 0.25 + 0.05

P( x >= 25 ) = 0.30

c) P( x > 0 ) = P( 10 ) + P( 25 ) + P( 50 ) = 0.30 + 0.25 + 0.05

P( x > 0 ) = 0.60

#3)

a) P( airline can accommodate everyone who shows up ) = P(95) +P(96)+P(97)+P(98)+P(99)+P(100 )

= 0.04+0.09+0.12+0.16+0.21+0.17 = 0.79

b) P( not all passengers can be accommodated ) = 1 - P( airline can accommodate everyone who shows up )

= 1 - 0.79 = 0.21

c) P (number 1 standby will be able to take the flight ) = P(95) +P(96)+P(97)+P(98)+P(99)

= 0.04+0.09+0.12+0.16+0.21

= 0.62

d) P( number 3 standby will be able to take the flight ) = P(95) +P(96)+P(97)

= 0.04+0.09+0.12

=0.25