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Sav [38]
3 years ago
6

A box contains three cards, labeled 1, 2, and 3. Two cards are chosen at random, with the first card being replaced before the s

econd card is drawn. Let X represent the number on the first card, and let Y represent the number on the second card.
A. Find the joint probability mass function of X and Y.
B. Find the marginal probability mass functions pX(x) and pY(y).
C. Find µX and µY.
D. Find µXY.
E. Find Cov(X,Y).
Mathematics
1 answer:
inysia [295]3 years ago
4 0

Answer:

Step-by-step explanation:

Given that a box ontains three cards, labeled 1, 2, and 3.

Two cards are chosen at random, with the first card being replaced before the second card is drawn.

Let X represent the number on the first card, and let Y represent the number on the second card.

Since drawing is done with replacement, we find that X and Y are independnet

A) Joint pdf of x and y are

     

y    x 0 1 2 3 Total

0 0.0625 0.0625 0.0625 0.0625 0.25

1 0.0625 0.0625 0.0625 0.0625 0.25

2 0.0625 0.0625 0.0625 0.0625 0.25

3 0.0625 0.0625 0.0625 0.0625 0.25

Total 0.25 0.25 0.25 0.25  

B) Pdf of X is

x        0     1       2       3

p    0.25  0.25 0.25 0.25

and Y also will have same pdf

C) \mu_x = 0.25(0+1+2+3) = 1.5\\\mu_y = 0.25(0+1+2+3) = 1.5

D) \mu_{xy } =E(x) E(y) = 2.25

E) Cov (x,y) =0 since X and Y are independent

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If the distribution is really (5.43,0.54)
defon

Answer:

0.7486 = 74.86% observations would be less than 5.79

Step-by-step explanation:

I suppose there was a small typing mistake, so i am going to use the distribution as N (5.43,0.54)

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

The general format of the normal distribution is:

N(mean, standard deviation)

Which means that:

\mu = 5.43, \sigma = 0.54

What proportion of observations would be less than 5.79?

This is the pvalue of Z when X = 5.79. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{5.79 - 5.43}{0.54}

Z = 0.67

Z = 0.67 has a pvalue of 0.7486

0.7486 = 74.86% observations would be less than 5.79

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Keep in mind

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As none of the 5,8,24 is a perfect square , nor addition of any two results in a perfect square, so square root of their addition or subtraction  or multiplication results in an  irrational number.

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