Question #100955
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.
1
Expert's answer
2020-01-06T05:35:16-0500

For a random variable,

XN(μ,σ2), z=XμσN(0,1)X \sim N(\mu,\sigma^2), \ z=\frac{X-\mu}{\sigma} \sim N(0,1) \newline

The given weight,


WN(60,92),Z=W609N(0,1)P(Z>a)=0.69W \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,a=-0.50,


P(Z>0.5)=P(Xμσ>W609)W=0.59+60=55.5P(Z>-0.5)=P(\frac{X-\mu}{\sigma}>\frac{W-60}{9}) \newline \because W=-0.5 \cdot 9+60=55.5



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