Question

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.908 g and a standard deviation of 0.319 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 46 cigarettes with a mean nicotine amount of 0.856 g.

Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly seleting 46 cigarettes with a mean of 0.856 g or less.
P(x-bar < 0.856 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 = = 0.908

standard deviation = = 0.319

n = 46

= = 0.908 and

= / n = 0.319 / 46 = 0.04703

P( < 0.856) =   P(( - ) / < (0.856 - 0.908) / 0.04703)

= P(z < -1.11)   Using standard normal table.   

Probability = 0.1335

No. The probability of obtaining this data is high enough to have been a chance occurrence.