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Nataliya [291]
3 years ago
12

The lengths of salamanders have a normal distribution with mean 15cm, and standard deviation 2cm. What length of salamander woul

d place it at the 80th percentile of salamander lengths?
Mathematics
1 answer:
Law Incorporation [45]3 years ago
4 0

Answer:

16.68 cm

Step-by-step explanation:

Given that :

Mean , m = 15

Standard deviation, s = 2

Length of salamander that would Place it at 80% of salamander length :

P(Z ≤ x) = 0.8

Zscore equivalent to 0.8 = 0.842

Using the relation :

Zscore = (x - m) /s

0.842 = (x - 15) / 2

1.684 = x - 15

Add 15 to both sides

1.684 + 15 = x - 15 + 15

16.684 = x

Hence, x = 16.68 cm

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A tobacco company claims that the amount of nicotene in its cigarettes is a random variable with mean 2.2 and standard deviation
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Answer:

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

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 random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 2.2, \sigma = 0.3, n = 100, s = \frac{0.3}{\sqrt{100}} = 0.03

What is the approximate probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true?

This is 1 subtracted by the pvalue of Z when X = 3.1. So

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

By the Central Limit Theorem

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

Z = \frac{3.1 - 2.2}{0.03}

Z = 30

Z = 30 has a pvalue of 1.

1 - 1 = 0

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

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