Question

The net weight in pounds of a packaged chemical herbicide is uniform for 49.62 < x < 50.04 pounds.
(a) Determine the mean of X. (Round your answer to 2 decimal places.) pounds
(b) Determine the variance of X. (Round your answer to 4 decimal places.) pounds2
(c) What is P(X < 50.1)

Answer 1

Answer:

a) $E(X)=\frac{50.64+49.62}{2}=50.13 pounds$

b) $Var(X)= \frac{(b-a)^2}{12} =\frac{(50.04-49.62)^2}{12}=0.0147$

c) $P(X<50.1)$

But we see that the distribution is defined ust between 49.62 and 50.04

And in order to find this we can use the CDF (Cumulative distribution function) given by:

$F(X) =\frac{x-a}{b-a}$

And if we replace we got:

$P(X<50.1) =P(X<50.04) +P(50.04\<X<50.1)= F(50.04)+0 = \frac{50.04-49.62}{50.04-49.62}+0= 1$

And makes sense since all the values are between 49.62 and 50.04

Step-by-step explanation:

For this case we define the random variable X ="net weigth in pounds of a packaged chemical herbicide" and the distribution for X is given by:

$X \sim U(a= 49.62, b=50.04)$

Part a

For the uniform distribution the expected value is given by $E(X) =\frac{a+b}{2}$ where X is the random variable, and a,b represent the limits for the distribution. If we apply this for our case we got:

$E(X)=\frac{50.64+49.62}{2}=50.13 pounds$

Part b

The variance for an uniform distribution is given by:

$Var(X)= \frac{(b-a)^2}{12}$

And if we replace we got:

$Var(X)= \frac{(b-a)^2}{12} =\frac{(50.04-49.62)^2}{12}=0.0147$

Part c

For this case we want to find this probability:

$P(X<50.1)$

But we see that the distribution is defined ust between 49.62 and 50.04

And in order to find this we can use the CDF (Cumulative distribution function) given by:

$F(X) =\frac{x-a}{b-a}$

And if we replace we got:

$P(X<50.1) =P(X<50.04) +P(50.04\<X<50.1)= F(50.04)+0 = \frac{50.04-49.62}{50.04-49.62}+0= 1$

And makes sense since all the values are between 49.62 and 50.04