Answer :
a.
[tex]\mathbb P(X>1000)=\mathbb P\left(\dfrac{X-1200}{100}>\dfrac{1000-1200}{100}\right)=\mathbb P(Z>-2)[/tex]
Since about 95% of a normal distribution falls within two standard deviations of the mean, that leaves 5% that lie without, with 2.5% lying to either side.
[tex]\mathbb P(Z>-2)=\mathbb P(|Z|<2)+\mathbb P(Z>2)=0.95+0.025=0.975[/tex]
b.
[tex]\mathbb P(1100<X<1300)=\mathbb P\left(\dfrac{1100-1200}{100}<\dfrac{X-1200}{100}<\dfrac{1300-1200}{100}\right)=\mathbb P(-1<Z<1)[/tex]
About 68% of a normal distribution lies within one standard deviation of the mean, so this probability is about 0.68.
c. You're looking for [tex]k[/tex] such that
[tex]\mathbb P(X>k)=0.10[/tex]
Since
[tex]\mathbb P(X>k)=\mathbb P\left(\dfrac{X-1200}{100}>\dfrac{k-1200}{100}\right)=\mathbb P(Z>k^*)=0.10[/tex]
occurs for [tex]k^*\approx1.2816[/tex], it follows that
[tex]\dfrac{k-1200}{100}=1.2816\implies k\approx1328[/tex]
So there's a probability of 0.10 for having a demand exceeding about 1328 pounds.
[tex]\mathbb P(X>1000)=\mathbb P\left(\dfrac{X-1200}{100}>\dfrac{1000-1200}{100}\right)=\mathbb P(Z>-2)[/tex]
Since about 95% of a normal distribution falls within two standard deviations of the mean, that leaves 5% that lie without, with 2.5% lying to either side.
[tex]\mathbb P(Z>-2)=\mathbb P(|Z|<2)+\mathbb P(Z>2)=0.95+0.025=0.975[/tex]
b.
[tex]\mathbb P(1100<X<1300)=\mathbb P\left(\dfrac{1100-1200}{100}<\dfrac{X-1200}{100}<\dfrac{1300-1200}{100}\right)=\mathbb P(-1<Z<1)[/tex]
About 68% of a normal distribution lies within one standard deviation of the mean, so this probability is about 0.68.
c. You're looking for [tex]k[/tex] such that
[tex]\mathbb P(X>k)=0.10[/tex]
Since
[tex]\mathbb P(X>k)=\mathbb P\left(\dfrac{X-1200}{100}>\dfrac{k-1200}{100}\right)=\mathbb P(Z>k^*)=0.10[/tex]
occurs for [tex]k^*\approx1.2816[/tex], it follows that
[tex]\dfrac{k-1200}{100}=1.2816\implies k\approx1328[/tex]
So there's a probability of 0.10 for having a demand exceeding about 1328 pounds.