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?

Answer :

Answer:

a) 0.3889

b) 0.5

c) 0.8333

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

Step-by-step 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:

[tex]P(X > x) = 1 - \frac{x - a}{b-a}[/tex]

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

[tex]P(c \leq X \leq d) = \frac{d - c}{b - a}[/tex]

The mean and the standard deviation are, respectively:

[tex]M = \frac{a+b}{2}[/tex]

[tex]S = \sqrt{\frac{b-a}^{2}{12}}[/tex]

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

This means that [tex]a = 160, b = 340[/tex]

a)

[tex]P(220 \leq X \leq 290) = \frac{290 - 220}{340 - 160} = 0.3889[/tex]

b)

[tex]P(160 \leq X \leq 250) = \frac{250 - 160}{340 - 160} = 0.5[/tex]

c)

[tex]P(X > 190) = 1 - \frac{190 - 160}{340 - 160} = 0.8333[/tex]

d)

[tex]M = \frac{160 + 340}{2} = 250[/tex]

[tex]S = \sqrt{\frac{340 - 160}^{2}{12}} = 51.96[/tex]

The mean is 250 and the standard deviation is 51.96.

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