The amounts with a mean a produces thes The supportim nicotine amou

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Q12
The amounts of nicotine in a certain brand of cigarette are normally distributed
with a mean of 0.957 g and a standard deviation of 0.291 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 0.888 g.
+
Assuming that the given mean and standard deviation have NOT changed, find
the probability of randomly selecting 46 cigarettes with a mean of 0.888 g or
less.
P(M < 0.888 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
ассepted.
Based on the result above, ¿is it valid to claim that the amount of nicotine is
lower? (Let's use a 5% cut-off for our definition of unusual.)
Yes. The probability of this data is unlikely to have occurred by chance alone?
No. The probability of obtaining this data is high enough to have been a chance
occurrence?|
11:46
Transcribed Image Text:A Calendar 6 Introduction to Soc b Answered: Tempera x WAMAP - Posts A WAMAP Assessme X min and max - Goo X E Untitled document x + A docs.google.com/document/d/1qYChxA37cmlGEWMpTrbRg31QUw9YBAOS06_8JGralKY/edit Untitled document * D O Saved to Drive A Share File Edit View Insert Format Tools Add-ons Help Last edit was seconds ago в IU A E = = = tE E - E - E E X 100% Normal text Times New. - 14 + 1 | 1 | 2 | 3 4 5 | 6 + 7 I Q12 The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.957 g and a standard deviation of 0.291 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 0.888 g. + Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly selecting 46 cigarettes with a mean of 0.888 g or less. P(M < 0.888 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 ассepted. Based on the result above, ¿is it valid to claim that the amount of nicotine is lower? (Let's use a 5% cut-off for our definition of unusual.) Yes. The probability of this data is unlikely to have occurred by chance alone? No. The probability of obtaining this data is high enough to have been a chance occurrence?| 11:46
Expert Solution
Step 1

A z-score represents the position of a data point in terms of its distance from the mean when measured in standard deviation units, i.e, how far the observation is from the mean in units of standard deviation.

The z of a variable (x) is the ratio of the difference of mean μ from the data point (x) by the standard deviation σ, i.e, z=x-μσ

 

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