Respuesta :
Answer:
a) [tex] X \sim Unif (a=148, b=164)[/tex]
b) [tex] E(X) = \frac{a+b}{2}= \frac{148+164}{2}=156 cm[/tex]
c) [tex] Var(X)= \frac{(b-a)^2}{12}= \frac{(164-148)^2}{12}= \frac{64}{3}= 21.33[/tex]
[tex] Sd(X)= \sqrt{21.33}= 4.6188[/tex]
d) [tex] P(X<155) = F(155)= \frac{155-148}{16}= \frac{7}{16}=0.4375[/tex]
e) [tex] P(155< X< 160)= F(160)-F(155)= \frac{160-148}{16} -\frac{155-148}{16}= 0.75-0.4375=0.3125[/tex]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:
[tex] X \sim Unif (a=148, b=164)[/tex]
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:
[tex] E(X) = \frac{a+b}{2}= \frac{148+164}{2}=156 cm[/tex]
c. Find the standard deviation.
First we need to calculate the variance given by:
[tex] Var(X)= \frac{(b-a)^2}{12}= \frac{(164-148)^2}{12}= \frac{64}{3}= 21.33[/tex]
[tex] Sd(X)= \sqrt{21.33}= 4.6188[/tex]
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:
[tex] f(x) =\frac{1}{b-a}= \frac{1}{164-148}= \frac{1}{16} , 148 \leq X \leq 164[/tex]
And the cumulative distribution function would be given by:
[tex] F(x) =\frac{x-148}{16} , 148 \leq X\leq 164[/tex]
We want to find this probability:
[tex] P(X<155) = F(155)= \frac{155-148}{16}= \frac{7}{16}=0.4375[/tex]
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:
[tex] P(155< X< 160)[/tex]
And we can use the cdf function and we have:
[tex]P(155<X<160) = F(160)-F(155)=\frac{160-148}{16} -\frac{155-148}{16}=0.75-0.4375=0.3125[/tex]
