ME 2014- Z score probability Normal-Distribution

Question

A nationalized bank has found that the daily balance available in its saving bank accounts follows a normal distribution with a mean of Rs. 500 and a standard deviation of Rs. 50. The percentage of savings account holders who maintain an average daily balance more than Rs. 500 is _______________.

 

Explain the calculation of the probability of Z score. Do GATE provide Z score table?

Comments

$P\left ( z\geq 500 \right ) =P\left ( z\geq \frac{500-\text Mean}{\text standard Deviation} \right )$

$P\left ( z\geq 500 \right ) = P\left ( z\geq \frac{500-\text 500}{\text 50} \right ) = P\left ( z\geq 0 \right )= 0.5$

Answer 1

probability $P(X>500)=0.5\  or\  50\%$

let variate be X,

$z=\frac{X-\mu}{\sigma}=\frac{X-500}{50}$ //converting into normalized form

$X=50z+500$

given, X>500, therefore $50z+500>500$ or $z>0$

probability, $P(z>0)=0.5$

therefore,  $P(X>500)=0.5\  or\  50\%$

PS: you should only know that $P(z\geq0)=P(z<0)=0.5$