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Olin [163]
3 years ago
10

Suppose we generate a random variable X in the following way. Forst we flip a fair coin. If the coin is heads, take X to have a

Uniform(0, 1). If the coin is tails, take X to have a Uniform(3, 4). (a) Find the mean of X. (b) Find the standard deviation of X.
Mathematics
1 answer:
otez555 [7]3 years ago
3 0

Answer:

(a) The mean of <em>X</em> is 2.

(b) The variance <em>X</em> is \frac{7}{3}<em>.</em>

Step-by-step explanation:

The random variable <em>X</em> defined as follows:

<em>X</em> = heads, then <em>X </em>follows Uniform (0, 1).

<em>X</em> = tails, then <em>X </em>follows Uniform (3, 4).

(a)

The expected value of a Uniform random variable U (a, b) is:

Mean=\frac{1}{2}(a+b)

Compute the expected value of <em>X </em>= heads as follows:

E(X=heads)=\frac{1}{2}(0+1)=\frac{1}{2}=0.50

Compute the expected value of <em>X </em>= tails as follows:

E(X=tails)=\frac{1}{2}(3+4)=\frac{7}{2}=3.50

Compute the mean of <em>X</em> as follows:

E(X)=\frac{E(X=heads)+E(X=tails)}{2} =\frac{0.50+3.50}{2}=\frac{4.00}{2}=2

Thus, the mean of <em>X</em> is 2.

(b)

Compute the variance of <em>X</em> as follows:

V(X)=\frac{1}{2}\int\limits^{1}_{0} {\frac{1}{1-0}[X-E(X)]^{2}} \, dx +\frac{1}{2}\int\limits^{4}_{3} {\frac{1}{4-3}[X-E(X)]^{2}} \, dx \\=\frac{1}{2}\int\limits^{1}_{0} {[x-2]^{2}} \, dx +\frac{1}{2}\int\limits^{4}_{3} {[x-2]^{2}} \, dx \\=\frac{1}{2}[\frac{(x-2)^{3}}{3}]^{1}_{0}+\frac{1}{2}[\frac{(x-2)^{3}}{3}]^{4}_{3}\\=(\frac{1}{2}\times\frac{7}{3})+(\frac{1}{2}\times\frac{7}{3})\\=\frac{7}{3}

Thus, the variance <em>X</em> is \frac{7}{3}<em>.</em>

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Answer:

Boxplot is constructed using quartiles and 5 No. summary means including minimum value of data, 1st quartile, 2nd quartile, 3rd quartile and maximum value of data.

As data already in increasing order, we find quartiles using the following formula,

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BoxPlot is attached with this ans.

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