Question

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.883 g and a standard deviation of 0.293 g. The company that produces these cigarettes claims that it has now reduced the amount of nicotine. In what range would you expect to find the middle 95% of amounts of nicotine in these cigarettes (assuming the mean has not changed)? Between and . If you were to draw samples of size 30 from this population, in what range would you expect to find the middle 95% of most average amounts of nicotine in the cigarettes in the sample? Between and . Enter your answers as numbers. Your answers should be accurate to 4 decimal places.

Answer 1

Solution,

Given that,

a) mean = = 0.883

standard deviation = = 0.293

Using standard normal table,

P( -z < Z < z) = 95%

= P(Z < z) - P(Z <-z ) = 0.95

= 2P(Z < z) - 1 = 0.95

= 2P(Z < z) = 1 + 0.95

= P(Z < z) = 1.95 / 2

= P(Z < z) = 0.975

= P(Z < 1.96 ) = 0.975

= z  ± 1.96

Using z-score formula,

x = z * +

x = -1.96 * 0.293 + 0.883

x = 0.3087

Using z-score formula,

x = z * +

x = 1.96 * 0.293 + 0.883

x = 1.4573

95% between 0.3087 and 1.4573

b) n = 30

= = 0.883

= / n = 0.293 / 30 = 0.05349

Using z-score formula  

= z * +

= -1.96 * 0.05349 + 0.883

= 0.7782

Using z-score formula  

= z * +

= 1.96 * 0.05349 + 0.883

= 0.9878

95% between 0.7782 and 0.9878