Question

Page >of 3 What Is the probability that the Aurora Borealis can be seen every day of this week a (b) What is the probability that the Aurora Borealis can be seen at least one day in this week? (c) What is the probability that the Aurora Borealis can be seen less than two days this week? (d) Find the mean, variance, and standard deviation for the number of days that the Aurora Borealis can be seen during this week. 2. There is a very large bin with many different types of dice. 50% of the dice in the bin are 6-sided dice, 25% are 10 sided dice, 20% are 20 sided dice, and 5% are the elusive 100 sided dice. Suppose we choose a die from this bin at random. Let X denote the number of sides of this randomly chosen die. (a) Construct a probability distribution table for X (b) Find the mean and standard deviation of X (c) What is the probability that the die will have more than 6 sides? (d) Given that the die has more than 6 sides, what is the probability that it will have exactly 100 sides? (e) Now for a challenge: Imagine we instead draw a random die like above, then roll the die to produce the random variable Y, whose value is the number shown on the rolled dic. Give the probability distribution table and mean for this new random variable 3. In the United States, 37% of the population has type Or blood. Consider taking a sample of size 12. Let X denote the number of persons in the sample with type O blood. Find (a) P(X 0). (b) /'(N1I) P<X6). (a) The mean and variauce of X

Question 2E

Answer 1

Question 2 (e)

Here draw a random die like above, then roll the die to produce the random variable Y.

SO here there are 50% , 6 - sided die

25%, 10 - sided die

20%, 20 - sided die and 5%, 100 - sided die.

so here Y is the number shown on the rolled die.

If we get a 6 - rolled die, then we will get numbers from 1 to 6.

P(Getting any number from 6 - rolled die) = 1/6

if we get a 10- rollled die, then we will get numbers from 1 to 10.

P(Getting any number from 10 - rolled die) = 1/10

if we get a 20- rollled die, then we will get numbers from 1 to 20.

P(Getting any number from 10 - rolled die) = 1/20

if we get a 100- rollled die, then we will get numbers from 1 to 100.

P(Getting any number from 100 - rolled die) = 1/100

so here Y has values from 1 to 100.

for Y = 1 to 6,

p(Y) = 1/6 * 0.5 + 1/10 * 0.25 + 1/20 * 0.2 + 1/100 * 0.05 = 0.118833

for Y = 7 to 10

p(Y) = 1/10 * 0.25 + 1/20 * 0.2 + 1/100 * 0.05 = 0.0355

for Y = 11 to 20

p(Y) = 1/20 * 0.2 + 1/100 * 0.05 = 0.0105

for Y = 21 to 100

p(Y) = 1/100 * 0.05 = 0.0005

so here

p(Y) = 0.118833 ; Y = 1- 6

= 0.0355 ; Y = 7-10

= 0.0105 ; Y = 11 - 20

= 0.0005 ; Y = 21 - 100

Here mean of Y = E[Y] = $\sum Yp(Y)$

= 3.5 * 0.5 + 5.5 * 0.25 + 10.5 * 0.20 + 50.5 * 0.05

= 7.75