Question

Sheets of aluminum from a supplier have a thickness that is normally distributed with a mean of 50 mm and a standard deviation of 4 mm (call this random variable X). Your company compresses the aluminum with a tool that is normally distributed with a mean of 20 mm and a standard deviation of 3 mm (call this random variable Y). You are interested in the random variable V = X – Y, the random variable V is the final aluminum thickness.

1. What is the probability that the outputted aluminum (that is, V), will be between 25 mm and 32 mm?

2. What is the probability that the outputted aluminum (that is, V), will be between 26.5 mm and 33.5 mm?

3. If the company had the choice between compressing aluminum to between 25-32 mm or 26.5-33.5 mm, then which is preferred (or neither)?

4. In one or two sentences, why is one preferred (if either) over the other [continuation of 3.3]; if neither are preferred, then why?

Answer 1

Given that ,

X N(50, 42)

And

Y N(20, 32)

Since ,

V= X-Y

...By additive property of normal distribution .

VN(50 20, 52 32) N(30, 5.832)

Where ,
д— 30тт )тm
  
5.83mm

and

We know that ,

N (0,1) Z =

1)Probability that the outputted aluminum (that is, V), will be between 25 mm and 32 mm is given by ,

P(25 V<32)

32- 25 P(25 V32) = P( 1- A

P(25 < V<32)- P(25-30 5,83 32 30 < Z < 5.83

P(25<V32) P-0.86< Z<0.34)

$P(25<V<32)=1-P(Z>0.34)-P(Z<-0.86)$     

..........Due to symmetry .

P(25<V<32) 1- P(Z> 0.34) P(Z > 0.86)

$P(25<V<32)=1-0.36693-0.19489$ .....................From normal probability table

$P(25<V<32)==0.43818=0.44$ .

2.

Probability that the outputted aluminum (that is, V), will be between 26.5 mm and 33.5 mm is given by ,

P(26.5 V < 33.5)

$P(26.5<V<33.5)=P(\frac{26.5-\mu }{\sigma }<\frac{V-\mu }{\sigma }<\frac{33.5-\mu }{\sigma })$

33.5 30. P(= P(26.5 V 33.5)-P 26.5-30 5.83 5.83

$P(26.5<V<33.5)=1-P(Z>0.6003)-P(Z<-0.6003)$     

$P(26.5<V<33.5)=1-P(Z>0.6003)-P(Z>0.6003)$ ..........Due to symmetry .

$P(26.5<V<33.5)=1-2P(>0.6003)=1-2*27425=0.4515$ .....................From normal probability table

$P(26.5<V<33.5)=0.4515=0.45$ .

3.

Company should had to prefer between 26.5 to 33.5 .

4. As the probability of outputted aluminium between 26.5 -33.5 is greater than probability of outputted aluminium between 25-32 .