Question

A certain industrial process yields a large number of steel cylinder whose lengths are distributed normally with mean 3.25 inches and standard deviation 0.05 inches. If two such cylinders are chosen at random and placed end to end what is the probability that their combined length is less than 6.60 inches? (round answer to two decimals)

Answer 1

Answer:

$Y = X_1 + X_2$

If we find the expected value of Y we got:

$E(Y) = E(X_1) + E(X_2) = 3.25 + 3.25 =6.5$

And for the variance we assume that we have independent random variables for X1 and X2 so then we have:

$Var(Y) = Var(X_1) + Var(X_2) = \sigma^2 + \sigma^2 = 2 \sigma^2 = 2*(0.05^2) = 0.005$

And the deviation $Sd(Y) = \sqrt{0.005}=0.0707$

So then our random variable Y have the following distribution:

$Y \sim N (\mu =6,5 , \sigma = 0.0707)$

And we want to find this probability:

$P(Y<6.60)$

And we can use the z score formula given by:

$z=\frac{y -\mu_y}{\sigma_y}$

And if we use the z score formula we got:

$P(Y<6.60)=P(Z<\frac{6.60-6.5}{0.0707}) = P(Z<1.414)=0.9213$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the lenghts of a population, and for this case we know the distribution for X is given by:

$X \sim N(3.25,0.05)$  

Where $\mu=3.25$ and $\sigma=0.05$

We slect two observations $X_1, X_2$

And let denote the combined lenght as:

$Y = X_1 + X_2$

If we find the expected value of Y we got:

$E(Y) = E(X_1) + E(X_2) = 3.25 + 3.25 =6.5$

And for the variance we assume that we have independent random variables for X1 and X2 so then we have:

$Var(Y) = Var(X_1) + Var(X_2) = \sigma^2 + \sigma^2 = 2 \sigma^2 = 2*(0.05^2) = 0.005$

And the deviation $Sd(Y) = \sqrt{0.005}=0.0707$

So then our random variable Y have the following distribution:

$Y \sim N (\mu =6,5 , \sigma = 0.0707)$

And we want to find this probability:

$P(Y<6.60)$

And we can use the z score formula given by:

$z=\frac{y -\mu_y}{\sigma_y}$

And if we use the z score formula we got:

$P(Y<6.60)=P(Z<\frac{6.60-6.5}{0.0707}) = P(Z<1.414)=0.9213$