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

Explanation:

Given: Mean :
`\mu =0`

Variance :
`\sigma^2=2.00`

⇒ Standard deviation :
`\sigma = √(2)`

The formula to calculate z is given by :-

`z=(x-\mu)/(\sigma)`

For x= -3

`z=(-3-0)/(√(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