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 proc…

Question

asked Oct 13, 2020 163k views

1 Answer

Emil Eriksson · Answer 1 · 2020-10-19T18:23:14+0000

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,
$P(X\leq 2.70)$ :

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

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