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 0.316 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 41 cigarettes with a mean nicotine amount of 0.893 g. Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly selecting 41 cigarettes with a mean of 0.893 g or less. P( ¯ x < 0.893 g) = Enter your answer as a number accurate to 4 decimal places. NOTE: Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted. Based on the result above, is it valid to claim that the amount of nicotine is lower? Note: we say a result is unusual if the probability of the event occurring or a more extreme event occurring is less than 0.05 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. Correct

Answer 1

Let x be the amounts of nicotine in a certain brand of cigarette.

x follows normal distribution with mean ( µ ) = 0.962 and standard deviation (σ) = 0.316

Sample size n = 41

According to sampling distribution of sample mean.

The sample mean (
T
) approximately follows normal distribution with mean $1596195267156_blob.png$ = $1596195267134_blob.png$ and standard deviation $1596195267150_blob.png$ =
n

Therefore $1596195267156_blob.png$ = 0.962 and $1596195267150_blob.png$ = $\frac{0.316}{\sqrt{41}}$ = 0.0494

We are asked to find P( < 0.893 )

X

= $P\left ( \frac{\bar{x}-\mu _{\bar{x}}}{\sigma _{\bar{x}}} <\frac{0.893-0.962}{0.0494}\right )$

= P( z ≤ -1.40)

= 0.0808 --- ( from z score table )

result is usual , because probability of the event is greater than 0.05

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