Question

Suppose that the measured voltage in a certain electric circuit has the normal distribution with mean 120 and standard deviation 2. If three independent measurements of the voltage are made, what is the probability that all three measurements will lie between 116 and 118?

Short answer

Probability that all three measurements will lie between 116 and 118 is\({p^3}\)

That is \({\left( {0.136} \right)^3}\)

Step-by-step solution

Given information

Voltage in the electric circuit has normal distribution with mean \(\left( \mu \right)\) is 120 and standard deviation \(\left( {{\sigma ^2}} \right)\) is 2.

Calculating the probability

If X is measurement having the specified normal distribution, and\(Z = \frac{{\left( {X - 120} \right)}}{2}\), then Z will have the standard normal distribution.

Therefore the probability that a particular measurement will lie in the given interval is

\(\begin{array}{c}p = {\rm P}\left( {116 < X < 118} \right)\\ = {\rm P}\left( { - 2 < X < - 1} \right)\end{array}\)

\(\begin{array}{c}p = {\rm P}\left( {1 < Z < 2} \right)\\ = \phi \left( 2 \right) - \phi \left( 1 \right)\\ = 0.1360\end{array}\)

As there are 3 measurements, therefore probability that all three measurement will lie in the interval is\({p^3}\)that is\({\left( {0.136} \right)^3}\).