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A random variable follows the continuous uniform distribution between 160 and 340. Calculate the following quantities for the distribution. ​a) ​P(220less than or equalsxless than or equals290​) ​b)​ P(160less than or equalsxless than or equals250​) ​c)​ P(xgreater than190​) ​d) What are the mean and standard deviation of this​ distribution?

User Otusweb
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1 Answer

5 votes

Answer:

a) 0.3889

b) 0.5

c) 0.8333

d) The mean is 250 and the standard deviation is 51.96.

Explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability of finding a value of X higher than x is:


P(X > x) = 1 - (x - a)/(b-a)

The probability of finding a value of X between c and d is:


P(c \leq X \leq d) = (d - c)/(b - a)

The mean and the standard deviation are, respectively:


M = (a+b)/(2)


S = \sqrt{\frac{b-a}^(2){12}}

A random variable follows the continuous uniform distribution between 160 and 340.

This means that
a = 160, b = 340

a)


P(220 \leq X \leq 290) = (290 - 220)/(340 - 160) = 0.3889

b)


P(160 \leq X \leq 250) = (250 - 160)/(340 - 160) = 0.5

c)


P(X > 190) = 1 - (190 - 160)/(340 - 160) = 0.8333

d)


M = (160 + 340)/(2) = 250


S = \sqrt{\frac{340 - 160}^(2){12}} = 51.96

The mean is 250 and the standard deviation is 51.96.

User Bobae
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