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madreJ [45]
3 years ago
11

A tobacco company claims that the amount of nicotine in its cigarettes is a randomvariable with mean 2.2 mg and standard deviati

on .3 mg. However, the sample mean nicotine content of100 randomly chosen cigarettes was 3.1 mg. What is the approximate probability that the sample meanwould have been as high or higher than 3.1 if the company’s claims were true?
Mathematics
1 answer:
Genrish500 [490]3 years ago
8 0

Answer:

0.0000

Unusual

Step-by-step explanation:

Given that a tobacco company claims that the amount of nicotine in its cigarettes is a random variable with mean 2.2 mg and standard deviation .3 mg.

i.e. population parameters are

\mu =2.2 \\s = 0.3

The approximate probability that the sample meanwould have been as high or higher than 3.1

=P(X\geq 3.1)\\=P(Z\geq \frac{3.1-2.2}{\frac{0.3}{\sqrt{100} } } )\\=P(Z\geq 30)\\

=0.0000

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ax+4y=6      +
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3/2

<em>Step One:</em> 1 3/8, transform the number into a improper fraction:

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Answer:

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Step-by-step explanation:

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Since,  \dfrac{\pi}{10}  between 0 and 2π, therefore,  the required coterminal angle is \dfrac{\pi}{10}.

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Step-by-step explanation:

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