Question

A worker at a landscape design center uses a machine to fill bags with potting soil. Assume that the quantity put in each bag follows the continuous uniform distribution with low and high filling weights of 10 pounds and 12 pounds, respectively. a. Calculate the expected value and the standard deviation of this distribution. (Do not round intermediate calculations. Round your "Standard deviation" answer to 4 decimal places.) Expected value Standard deviation b. Find the probability that the weight of a randomly selected bag is no more than 11 pounds. (Round your answer to 2 decimal places.) Probability c. Find the probability that the weight of a randomly selected bag is at least 10.5 pounds. (Round your answer to 2 decimal places.)

Answer 1

Answer:

a) The expected value is given by:

$E(X) =\frac{a+b}{2}= \frac{10+12}{2}= 11$

The variance is given by:

$Var(X) = \frac{(b-a)^2}{12}= \frac{(12-10)^2}{12}= 0.3333$

And the standard deviation is just the square root of the variance and we got:

$Sd(x) = \sqrt{0.3333}= 0.5774$

b) $P(X$

And we can use the cumulative distribution function given by:

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

And using this function we got:

$P(X$

c) $P(X>10.5)$

And using the complement rule and the cumulative distribution function we got:

$P(X>10.5)= 1-P(X$

Step-by-step explanation:

For this case we define the random variable X=quantity put in each bag , and we know that the distribution for X is given by:

$X \sim Unif (a= 10, b =12)$

Part a

The expected value is given by:

$E(X) =\frac{a+b}{2}= \frac{10+12}{2}= 11$

The variance is given by:

$Var(X) = \frac{(b-a)^2}{12}= \frac{(12-10)^2}{12}= 0.3333$

And the standard deviation is just the square root of the variance and we got:

$Sd(x) = \sqrt{0.3333}= 0.5774$

Part b

For this case we want this probability:

$P(X$

And we can use the cumulative distribution function given by:

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

And using this function we got:

$P(X$

Part c

For this case we want this probability:

$P(X>10.5)$

And using the complement rule and the cumulative distribution function we got:

$P(X>10.5)= 1-P(X$