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

Answer:

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, $P(X\leq 2.70)$:

$z = \frac{x - \mu}{\sigma} = \frac{2.70 - 2.501}{0.113} = 1.761$

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:

$P(X\geq 2.70) = \frac{No.\ of\ readings\ expected\ to\ be\ greater\ than\ 2.70\ V}{Total\ no.\ of\ readings}$

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

No. of readings expected to be greater than 2.70 V = $0.0392\times 60 = 2.352$ ≈ 2