An article presents voltage measurements for a sample of 66 industrial networks in estonia. assume the rated voltage for these networks is 232 v. the sample mean voltage was 231.7 v with a standard deviation of 2.19 v. let μμ represent the population mean voltage for these networks.

a sample of 29 network will have a voltage of 232V

Step-by-step 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

step-by-step explanation:

where is the questions

b

step-by-step explanation: