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AnnyKZ [126]
3 years ago
15

A sensitive measuring device is calibrated so that errors in the measurements it provides are normally distributed with mean 0 a

nd variance 2.00. Find the probability that a given error will be between -3 and 3.
Mathematics
1 answer:
Artemon [7]3 years ago
3 0

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

For x= 3

The P Value =P(z

\text{Now, }P(-3

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

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