Answer to Question #100955 in Statistics and Probability for shermine

Question

The weight of students follows a Normal distribution with a mean weight of 60 kg and a standard deviation of 9 kg. 69% of students weigh more than W kg. Find W.

Answer 1

For a random variable,

$X \sim N(\mu,\sigma^2), \ z=\frac{X-\mu}{\sigma} \sim N(0,1) \newline$

The given weight,

$W \sim N(60,9^2), Z=\frac{W-60}{9} \sim N(0,1) \newline P(Z>a)=0.69$

From the normal distribution Z table,

$a=-0.50,$

$P(Z>-0.5)=P(\frac{X-\mu}{\sigma}>\frac{W-60}{9}) \newline \because W=-0.5 \cdot 9+60=55.5$