Answer:
The answer is 15.68 in
Explanation:
X~N(μ=14.2; σ=0.9)
[tex]P(X\leq a)=0.95[/tex]
I want to find "a", so:
[tex]P(0\leq X\leq a)=P(\frac{0-\mu}{\sigma} \leq \frac{X-\mu}{\sigma}\leq \frac{a-\mu}{\sigma}), Z= \frac{X-\mu}{\sigma}[/tex]
Z~N(0; 1)
[tex]P(0\leq X\leq a)=P(\frac{0-14.2}{0.9} \leq Z}\leq \frac{a-14.2}{0.9})=\phi(\frac{a-14.2}{0.9})-\phi(-15.78)=0.95[/tex]
Φ(-15.78)≅0
[tex]\phi(\frac{a-14.2}{0.9})=0.95[/tex]
Based on the Standard Normal table
[tex]\phi(Z_{a} )=0.95[/tex] if [tex]Z_{a}=1.64485[/tex]
(a-14.2)/0.9=1.64485, then a=15.68