Assignment Overview:
Given, Y=the number of tests in which the two defectives can be found.
P(Y)= the probability of the number of tests in which the two defectives can be found.
The total number of tests which can be done=
Y=2, the total number of ways in which we can find the defectives when only two tests are done==6
The number of odds in favor==1.
Hence, P(Y=2) =
Similarly when Y=3,
Similarly when Y=4,
The probability of a cell dying is 0.1 and the probability of the same cell splitting is 0.9 so, the two events are independent and mutually exclusive.
Thus, the probability distribution of finding the number of cells in new generation is given by the Bernoulli distribution.
P (n)=
3.11This a binomial distribution with the probability given as:
Where, n is the number of trails and r the number of successes. Here, and.
Similarly for the variable Y, and.
Similarly for Z=X+Y, and.
3.12
=2
=0.641
==5
So, =5-1=4
==5-4=1
3.15. Let C be the premium of the company, then the profit of the company would be C-15 if A does not occur and C-15-1000 if A does occur.
So, the expected profit=
Thence the equation:
Hence the valve of the policy should be $85.
3.22 The single fair coin is tossed;hence they have the following distribution.
| y | 1 | 2 | 3 | 4 | 5 | 6 |
| p(y) |
=3.5
=15.16
So, ==230-12.25=217.75
3.37 a. The number of students who took the SAT forms a population for the binomial distribution but the size cannot be determined from the given data.
b. The scores of the 100 students from a sample for the binomial distribution with the size n=100
c. The number of students in the sample who scored above average can be found out from the binomial distribution but here n=100 but p cannot be determined from the given data.
d. The amount of time required to complete the SAT cannot be found using binomial distribution.
e. The number of female high school grads cannot be found using binomial distribution.
3.40 Given p=0.8 for the binomial distribution.
a ==0.196-0.087=0.109
b. =1-=1-0.003=0.997
c. ==0.931-0.196=0.735
d. =0.589
The table of binomial distribution with n=20 was used in the above problem.
3.44
a. Given p=0.8, ==0.327
b. Given p=0.6, ==0.1296
c. Given p=0.3,=0.528
3.52 In the following binomial probability the probability p of tasters is 0.7.
a. = 1-=1-0.965=0.035
b. 0.584
The table of binomial distribution with n=20 was used in the above problem.
3.121
a. ==0.09
b. =1-=1-0.947=0.053
c. ==0.857
d. ==0.053+1-=1.053-0.677=0.376
The table 3 of appendix 3 was used to calculate the probabilities from b to d.
3.124 Given probability.
Hence, from the table it can be found that for the given Poisson distribution.
Thus, =1-)
=1- 0.895=0.105
3.127 The average of typing errors.
We need to calculate the P(X), where X denotes the number of typing errors. The appendix 3 table 3 gives P(X) =0.629.
Thus, the probability that a page is not retyped is 0.629
3.135 The probability of success of the salesman on single contract is 0.03.
Clearly this is a binomial distribution of which we need to calculate the probability of making at least one sale in 100 prospects.
=1-0.1945=0.8055
4.3
a. The following graph shows the Bernoulli ‘s distribution with p(0)=1-p=q, p(1)=p.
b. The Bernoulli’s distribution clearly verifies the theorem 4.1 as
| y<0 | y>1 | ||
| F(y) | 0 | q | 1 |
4.7 a. The probabilities are not equal because the probabilities of a discrete binomial distribution would depend on the endpoints.
b. Similar the case again as part a,the probabilities are not equal because the probabilities of a discrete binomial distribution would depend on the endpoints.
c. No, the claim is true for a continuous distribution. The values of the integrals do not depend whether the endpoints are included or excluded, so it holds true for the continuous distributions.
4.9 a. This is a discrete random variable because it takes different values in different intervals.
b. The values of Y are 2, 2.5,4, 5.5,6 ,7.
c.
| 2 | ||||||
| 1/8 | 1/16 | 5/16 | 1/8 | 1/16 | 5/16 |
d.Median and the fifth percentile of the given distribution function would be equal to 4.
4.13
a.
| y<0 | y>1.5 | |||
| F(y) | 0 | 1 |
b. ==0.125
c. ==0.375+0.2=0.575
4.21 The mean of the density function is calculated, as follows:
== from.
=
=
=
=from.
=0.026
4.29
As above we have the mean as follows:
= Where
= 60
= Where
= 0.33
4.33The probability density function of the solar radiation is given as:
=
Hence, the mean expected daily solar radiation is given as:
= from
=4
4.45 a.
b.
4.49.
4.50.
a. P= (25/500)
b. P= (25/500)
c. The probability would be less than 0.5
4.51 (The interval reduces from 20 to 15 because of the condition that the cycle time exceeds 55 min.)
4.58
a. =0.3849
b. =0.3159
c.=0.4406-0.1179=0.3227
4.59
a. =0.5, Clearly =0
b. =0.8643 or =0.8643-0.5=0.3643 so, =1.1
c. =0.90 or =0.90-0.5=0.40 so, =1.28
d. =0.99 or =0.99-0.5=0.49 so, =2.33
4.63 Clearly the Z==1
So the proportion of bottles having more than 17 ounces=0.3413
4.71 a. =-2
Hence, the probability of wires having z-values greater than -2=0.4772
=2
Hence, the probability of wires having z-values less than 2=0.4472
Thus the total probability of having the wire resistances between 0.13-0.14 ohms is 0.9544
b. The probability is equal to as all the probability is independent.
4.75 The probability of 1% tells us that the z value for which the cups overflow is 0.25. Hence, if the cups exceed 0.075 ounce over the mean the cups, the cups would overflow in 1% of the time.
LI53
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