Question

The resistances of carbon resistors of 1500 (in ohms) nominal value are normally distributed with mean μ = 1500 and σ = 200. Within plus and minus what amount around the mean do 80 percent of resistances fall?

Answer 1

Solution:-

Given that,

mean = = 1500

standard deviation = = 200

Using standard normal table,

P( -z < Z < z) = 80%

= P(Z < z) - P(Z <-z ) = 0.80

= 2P(Z < z) - 1 = 0.80

= 2P(Z < z) = 1 + 0.80

= P(Z < z) = 1.80 / 2

= P(Z < z) = 0.90

= P(Z < 1.28 ) = 0.90

= z  ± 1.28

Using z-score formula,

x = z * +

x = -1.28 * 200 + 1500

x = 1244

Using z-score formula,

x = z * +

x = 1.28 * 200 + 1500

x = 1756

The 80% of resistances fall between 1244 and 1756