Question

A student takes 60 voltages readings across a resistor and finds a mean voltage of 2.501V with a sample standard deviation of 0.113V. Assuming that errors are due to random processes, how many of the readings are expected to be greater than 2.70V?

Answer 1

There are 2 expected readings greater than 2.70 V

Solution:

As per the question:

Total no. of readings, n = 60 V

Mean of the voltage,
`\mu = 2.501 V`

standard deviation,
`\sigma = 0.113 V`

Now, to find the no. of readings greater than 2.70 V, we find:

The probability of the readings less than 2.70 V,
`P(X\leq 2.70)`:

`z = (x - \mu)/(\sigma) = (2.70 - 2.501)/(0.113) = 1.761`

Now, from the Probability table of standard normal distribution:

`P(z\leq 1.761) = 0.9608`

Now,

`P(X\geq 2.70) = 1 - P(X\leq 2.70) = 1 - 0.9608 = 0.0392 = 3.92%`

Now, for the expected no. of readings greater than 2.70 V:

`P(X\geq 2.70) = (No.\ of\ readings\ expected\ to\ be\ greater\ than\ 2.70\ V)/(Total\ no.\ of\ readings)`

No. of readings expected to be greater than 2.70 V =
`P(X\geq 2.70)* Total\ no.\ of\ readings`

No. of readings expected to be greater than 2.70 V =
`0.0392* 60 = 2.352` ≈ 2