Question

A boardwalk is to be erected along an ocean beach. The specifications for the boardwalk require a width of at least 155 cm., but the pre-cut pieces of lumber supplied for the boardwalk are uniformly distributed in length between 148 cm. and 164 cm.

Answer 1

Answer:

a) $X \sim Unif (a=148, b=164)$

b) $E(X) = \frac{a+b}{2}= \frac{148+164}{2}=156 cm$

c) $Var(X)= \frac{(b-a)^2}{12}= \frac{(164-148)^2}{12}= \frac{64}{3}= 21.33$

$Sd(X)= \sqrt{21.33}= 4.6188$

d) $P(X$

e) $P(155< X< 160)= F(160)-F(155)= \frac{160-148}{16} -\frac{155-148}{16}= 0.75-0.4375=0.3125$Step-by-step explanation:

Assuming the following questions:

a. The distribution is X Use whole numbers.

Let X the random variable that represent the "lenght of pieces" used for the construction. We know that X follows an uniform distribution given by:

$X \sim Unif (a=148, b=164)$

b. The average length of the pieces of lumber is cm. Use whole numbers.

For this case the expected value for the distribution is given by:

$E(X) = \frac{a+b}{2}= \frac{148+164}{2}=156 cm$

c. Find the standard deviation.

First we need to calculate the variance given by:

$Var(X)= \frac{(b-a)^2}{12}= \frac{(164-148)^2}{12}= \frac{64}{3}= 21.33$

$Sd(X)= \sqrt{21.33}= 4.6188$

d. What's the probability that a randomly chosen piece of lumbar will be less than the required lenght?

For this case we can use the density function for X given by:

$f(x) =\frac{1}{b-a}= \frac{1}{164-148}= \frac{1}{16} , 148 \leq X \leq 164$

And the cumulative distribution function would be given by:

$F(x) =\frac{x-148}{16} , 148 \leq X\leq 164$

We want to find this probability:

$P(X$

e. Find the probability that a randomly chosen piece of lumber will be between 155 and 160 cm long?

For this case we want this probability:

$P(155< X< 160)$

And we can use the cdf function and we have:

$P(155$