Question

In the production of protective film for solar panels, precision is very important. The film is known
to have a thickness that is normally distributed with a standard deviation of σ = 5 microns. How
large a sample of film should be employed if we desire the 99 percent confidence interval for µ to
have a width of at most 2 microns? Sample size must be an integer. To guarantee that the width
of the confidence interval is not more than 2 microns, round your solution up to the nearest integer. thanks

Answer 1

Solution :

Given that,

standard deviation = $\sigma$ = 5

width = 2

E = width / 2 = E = 2 / 2 = 1

E = 1

margin of error = E = 1

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

$\alpha$ / 2 = 0.01 / 2 = 0.005

Z_$\alpha$/2 = Z_0.005 = 2.576

Sample size = n = ((Z_$\alpha$/2 * $\sigma$ ) / E)²

= ((2.576 * 5) / 1)²

= 165.8

Sample size = 166