Question

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.893 g and a standard deviation of 0.306 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 44 cigarettes with a mean nicotine amount of 0.819 g.

Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly selecting 44 cigarettes with a mean of 0.819 g or less.
P(M < 0.819 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?

• No. The probability of obtaining this data is high enough to have been a chance occurrence.
• Yes. The probability of this data is unlikely to have occurred by chance alone.

Answer 1

Solution :

Given that ,

mean = = 0.893

standard deviation = = 0.306

n = 44

= = 0.893 and

= / n = 0.306 / 44 = 0.0461

P(M < 0.819) =   P((M - ) / < (0.819 - 0.893) / 0.0461)

= P(z < -1.61)   Using standard normal table.   

Probability = 0.0536

Yes. The probability of this data is unlikely to have occurred by chance alone.