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. a. the distribution is x ~ ( , ). use whole numbers. b. the average length of the pieces of lumber is cm. use whole numbers. c. find the standard deviation. round to 2 decimals. d. what's the probability that a randomly chosen piece of lumber will be less than the required length? round to 4 decimals. e. find the probabililty that a randomly chosen piece of lumber will be between 155 and 160 cm long? round to 4 decimals. f. find the third quartile. cm.use whole numbers.

Answer 1

Here we find the answer Mean, Standard Deviation ,Variance, and Probability of the given equation:

1. $X\sim Unif\left(a=148,b=164\right)$

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

   

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

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

   4. $P\left(X < 155\right)=F\left(155\right)=\frac{\left(155-148\right)}{16}=\frac{7}{16}=0.4375$

    5. $P\left(155 < X < 160\right)=F\left(160\right)-F\left(155\right)=\frac{160-148}{16}-\frac{155-148}{16}=0.75-0.4375=0.3125$

     

What is Variance (Var)?

Variance is really a way of measuring dispersion that takes under consideration the scatter of all the information points in a data set.

What is Standard Deviation (Sd)?

Standard deviation (or σ) is a measure of just how dispersed the info is within relation to the suggest.

Explanation for the solution:

By considering the following question:

1) The distribution is X Use whole numbers.

Let X the variable this is certainly arbitrary express the "length of pieces" used for the building. We realize that X follows a circulation that is uniform by:

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

2) 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:

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

3) Standard deviation will be:

First, we need to calculate the variance given by:

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

Now, we calculate the standard deviation:

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

4) What's the probability that a randomly chosen piece of lumbar will be less than the required length?

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

$f(x)=\frac{1}{(b-a)}=\frac{1}{164-148}=\frac{1}{16},\ 148\le X\le164$

And the cumulative distribution function would be given by:

$F(x)=\frac{x-148}{16}\ ,\ 148\le X\le164$

We want to find this probability:

$P\left(X < 155\right)=F\left(155\right)=\frac{\left(155-148\right)}{16}=\frac{7}{16}=0.4375$

5)  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:

$P\left(155 < X < 160\right)$

And we can use the cumulative distribution function and we have:

$P\left(155 < X < 160\right)=F\left(160\right)-F\left(155\right)=\frac{160-148}{16}-\frac{155-148}{16}=0.75-0.4375=0.3125$

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