Question

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.894 g and a standard deviation of 0.326 g. The company that produces these cigarettes claims that it has now reduced the amount of nicotine. The supporting evidence consists of a sample of 32 cigarettes with a mean nicotine amount of 0.825 g. Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly seleting 32 cigarettes with a mean of 0.825 g or less. P(x-bar < 0.825 g) = Enter your answer as a number accurate to 4 decimal places. Based on the result above, is it valid to claim that the amount of nicotine is lower?

Answer 1

Solution :

Given that ,

mean = = 0.894

standard deviation = = 0.326

n = 32

=   = 0.894

= / n = 0.326/ 32 = 0.0576

P( < 0.825) = P(( - ) / < (0.825 - 0.894) / 0.0576 )

= P(z < -1.20)

Using z table

= 0.1151