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emmasim [6.3K]
2 years ago
7

PLEASE I NEED HELP WITH THIS SO I CAN PASS

Mathematics
2 answers:
Neko [114]2 years ago
6 0

Answer:

the answer is------------ 1/4.4.4

torisob [31]2 years ago
5 0
Guess ur not passing today
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The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.895 g and a standard deviation
kvasek [131]

Answer:

3.67% probability of randomly seleting 37 cigarettes with a mean of 0.809 g or less.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 0.895, \sigma = 0.292, n = 37, s = \frac{0.292}{\sqrt{37}} = 0.048

Find the probability of randomly seleting 37 cigarettes with a mean of 0.809 g or less.

This is the pvalue of Z when X = 0.809.

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{0.809 - 0.895}{0.048}

Z = -1.79

Z = -1.79 has a pvalue of 0.0367

3.67% probability of randomly seleting 37 cigarettes with a mean of 0.809 g or less.

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Step-by-step explanation:

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