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Deffense [45]
3 years ago
5

A particular type of mouse's weights are normally distributed, with a mean of 359 grams and a standard deviation of 33 grams. If

you pick one mouse at random, find the following: (round all probabilities to four decimal places)
a) What is the probability that the mouse weighs less than 405 grams?
b) What is the probability that the mouse weighs more than 461 grams?
c) What is the probability that the mouse weighs between 406 and 461 grams?
d) Is it unlikely that a randomly chosen mouse would weigh less than 405 grams?
i. No, the likelihood exceeds 50%
ii. No, the likelihood exceeds 5%
iii. Yes, the likelihood is less than 50%
iv. Yes, the likelihood is less than 5%
e) What is the cutoff for the heaviest 10% of this type of mouse?
Mathematics
1 answer:
igor_vitrenko [27]3 years ago
5 0

Answer:

Following are the responses to the given points:

Step-by-step explanation:

Given:

\mu = 359 \ \ \ \ \ \sigma = 33\\\\

Using formula:

\to P(X = x) = P\left ( z < \frac{x-\mu }{\sigma } \right )

For point a:

\to x= 406\\\\P(X < 406 ) = P\left ( z < \frac{406-359 }{33 } \right )\\\\

                   = P\left ( z

For point b:

\to x= 461\\\\P(X > 461 ) = P\left ( z > \frac{461 -359 }{33 } \right )\\\\

                   = 1 - P\left ( z < 3.0909 \right )\\\\= 1 - 0.9990\\\\= 0.0001

For point c:

\to x= 406\ \ and \ \  461\\\\P(406

                              = 0.9990 - 0.9222\\\\= 0.0768\\\\

For point d:

P(X < 406) = 0.9222 = 92.22 \%\  that \ is \ greater \ than\ 50\%

Hence, the correct choice is "i".

For point e:

\to P(X > z) = 0.1\\\\\to 1 - P(X < z) = 0.1\\\\\to P(\frac{x-\mu }{\sigma }) = 0.9\\\\\to \frac{x-359}{33 } = 1.28\\\\ \to x = 401.24\\\\

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Step-by-step explanation:

To solve this question, we have 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, which is also called standard error s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 13.1, \sigma = 4.1, n = 25, s = \frac{4.1}{\sqrt{25}} = 0.82

What is the probability that their mean length is less than 11.1 inches

This is the pvalue of Z when X = 11.1. So

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

By the Central Limit Theorem

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Z = -2.44 has a pvalue of 0.0073.

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