Answer:
a sample of 29 network will have a voltage of 232V
Explanation:
N = 66
variate = 232 V.
mean voltage = 231.7 V
standard deviation = 2.19 V.
The z-value given by = (variate -mean)/ standard deviation
= (232-231.7)/2.19 =
z = 0.3/2.19 = 0.137
From of normal distribution table, the z of 0.137 found between the area of z=0 and z=0.5 is given as 0.0557
Thus the area to the right of the z=0.0557 ordinate is 0.5000−0.0557=0.4443 = 44.43%
thus, this is the probability of Network voltage = 44.43%.
for sample of 66 network, it is likely that 66×0.4443 = 29.32, i.e.
a sample of 29 network will have a voltage of 232V