(a) (i) This is a case of Normal distribution. A normal distribution is one
suited to describe continuous random variables. It is appropriate in many situations where measurement is involved, such as weight, volume, heights of people and the life times of electrical goods. The mean and the standard deviation are also given. Unlike Normal distribution, the PQ Binomial distribution has its mean given as µ = np.
Where P – probability of success. n – number of trials.
Variance for the Binomial distribution d2 = npq.
Where q – is the probability of failure while p and n are as explained above.
The poisson distribution has a constant mean “? ” per unit interval with the variance equal to the mean “x? ”
(ii) For normally distributed function, N(µ,(), the standard values
x(µ
Z =
(
Where ( - is the standard deviation. µ - is the mean. x - is the observation.
600 (800
In this case, Z = = - 1.25
160
And from the normal tables:
-1.25 µ
The area under the curve represents the probability or the proportion of events. In this case, the percentage components to be replaced is 10.3470.
From the normal table, an area of 0.49 under the table corresponds to a Z value of 2.33. But since the Bulb with the shortest life is required, this is below the mean and must take a negative value. The value to be used as the guaranteed lifetime for bulbs, denoted as x, can be obtained from the expression for the Z value of -2.33 as follows:
x (µ
( Z =
d
x (800
( -2.33 =
160
( -2.33 x 160 = x – 800
( 800 – 2.33 x 160 = x
( 427.2
The guaranteed life time should be 427 Hours.
(iv) When all possible samples of size n are drawn from a population mean with a mean µand a standard deviation d, then the means of the samples have a probability distribution known as the sampling distribution of means with mean equal to the population mean µ, and a standard
d deviation = where dis the population standard deviation from which
the samples are drawn.
The standard deviation of the sampling distribution of the mean is known as the standard error of the mean. In this case a sample of size
25 will have a mean x(800Hours and a standard error of the mean =
160 160
( 32.
Then using the standard values, Z = ( 1.56 or 2
0 2
From the Normal table a Z value of 2 corresponds to an area = 0.4772, But we require the probability that it exceeds 850 Hours. This is the area to the Right of Z = 2 which is equal to 0.5 – 0.4772 = 0.023.
,b)(i) This is a case of Binomial distribution. This is because the interest is success or failure and the situation deals with discrete variables and not continuous variables. The probability of success here, is P = 0.15, getting a perishable product. The probability of failure is then q = 1.00 – 0.15 = 0.85.
ii)The commodity will be purchased if not more than 2 perishable items are found in sample of 10. The probability of not more than 2 perishable can be written as:
Probability of 0 perishable, 1 perishable or 2 perishable. The probability of 0 perishable, 1 perishable and 2 perishable are obtained using the binomial expression as follows:
The binomial probability expression states that the probability of event (x) written P(x) = nCxpxqn – x.
Thus:
P(0) = 10C0(0.15)0(0.85)10 = (0.15)0 (0.85)10
!
=
0.1968
P(1) = 10C1(0.15)1(.85)9
= 0.3474
P(2) = 10C2(0.15)2(0.85)8
= 0.2759
0.820
Recall: nCrPr qn-r =
n!
r n(r
P q
(n ( r)! r!
The probability of buying the consignment is 0.820
c)(i) This is a case of Poisson distribution. It deals with discrete variables and the probability of a given number of events happening within a specified time interval. The poisson distribution also deals with success or failure but within a specified time interval. This is poisson distribution with a mean “? ” of 2 per 10 seconds or 0.2 per second.
ii)The probability of an event occurring is computed using the following formula for a poisson distribution.
x (?
? e
P(x) =
x!
The probability of more than 3 cars can be expressed as follows:
P(x>3) = 1 – probability of 0, 1, 2, and 3 cars passing. Using the Poisson formula, we get:
P(0)
=
(1.2) 0 e(0.2
(
0!
0.818
P(1)
=
(0.2) 1 e(0.2
(
1!
0.164
P(2)
=
(0.2) 2 .e(0.2
=
2!
0.016
P(3)
=
(0.2) 3 e(0.2
=
3!
0.0011
0.9991
(Probability of more than 3 cars = 1 – 0.9991
= 0.0009
Moraa orina answered the question on April 2, 2018 at 18:23
-
a)Explain briefly the application of calculus to economic models, especially in the context of maximizing contribution and minimizing costs.
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a)Explain briefly the application of calculus to economic models, especially in the context of maximizing contribution and minimizing costs.
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Company A can restrict the weekly supply of its product to company B, but cannot raise the unit price above Sh. 200. The two companies are considering whether to merge together into a single company.
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