Answer:
(a)[tex]0.15731[/tex]
(b)0.02275
Step-by-step explanation:
We are given that
Mean=0
Standard deviation=0.5 g
True weight of a sample=166 g
Let X denote the normal random variable with mean =166+0=166
(a)
P(166.5<X<167.5)
=[tex]P(\frac{166.5-166}{0.5}<\frac{X-\mu}{\sigma}<\frac{167.5-166}{0.5})[/tex]
=[tex]P(1<Z<3)[/tex]
=[tex]P(Z<3)-P(Z<1)[/tex]
[tex]=0.99865-0.84134[/tex]
[tex]=0.15731[/tex]
(b)
[tex]P(X>167)=P(Z>\frac{167-166}{0.5})[/tex]
[tex]=P(Z>2)[/tex]
[tex]=1-P(Z<2)[/tex]
[tex]=1-0.97725[/tex]
[tex]=0.02275[/tex]