Question

A sensitive measuring device is calibrated so that errors in the measurements it provides are normally distributed with mean 0 and variance 2.00. Find the probability that a given error will be between -3 and 3.

Answer 1

Answer: 0.9660

Step-by-step explanation:

Given: Mean : $\mu =0$

Variance : $\sigma^2=2.00$

⇒ Standard deviation : $\sigma = \sqrt{2}$

The formula to calculate z is given by :-

$z=\dfrac{x-\mu}{\sigma}$

For x= -3

$z=\dfrac{-3-0}{\sqrt{2}}=-2.12132034356\approx-2.12$

The P Value =$P(z<-2.12)=0.017003$

For x= 3

The P Value =$P(z<2.12)=0.9829969$

$\text{Now, }P(-3<X<3)=P(X<3)-P(X<-3)\\\\=P(z<2.12)-P(z<-2.12)\\\\=0.9829969-0.017003=0.9659939\approx 0.9660$

Hence, the probability that a given error will be between -3 and 3=0.9660