Question

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.962 g and a standard deviation of 0.313 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 48 cigarettes with a mean nicotine amount of 0.899 g.

Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly seleting 48 cigarettes with a mean of 0.899 g or less.  
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 (greater than a 5% chance) to have been a chance occurrence.
• Yes. The probability of this data is unlikely (less than a 5% chance) to have occurred by chance alone.

Answer 1

Solution :

Given that ,

mean = = 0.962 g

standard deviation = = 0.313 g

n = 48

= = 0.962 g

= / n = 0.313 / 48 = 0.0452

P( 0.899 ) = P(( - ) / (0.899 - 0.962) / 0.0452 )

= P(z -1.39)

Using z table

= 0.0823

No. The probability of obtaining this data is high enough (greater than a 5% chance) to have been a chance occurrence.