Answered: a) Find the probability of existence of a random sample (size 100) weight within + 5 lb. of population mean weight. b) For the same sample size, find the…

MATLAB: An Introduction with Applications

6th Edition

ISBN: 9781119256830

Author: Amos Gilat

Publisher: John Wiley & Sons Inc

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Estimation

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.

Question

Wild boar population weight has standard deviation of 50 lb.

a) Find the probability of existence of a random sample (size 100) weight within + 5 lb. of population mean weight.

b) For the same sample size, find the probability of the weight within + 10 lb. of population mean weight.

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Expert Solution

Step 1

Wild boar population weight has standard deviation of 50 lb. That is, $σ = 50$ .

The probability of existence of a random sample (size 100) weight within $\pm$ 5 lb. of population mean weight is,

$\begin{array}{rcl} P (- 5 < \bar{x} - μ < 5) & = & P (\frac{- 5}{(\frac{50}{\sqrt{100}})} < \frac{\bar{x} - μ}{(\frac{σ}{\sqrt{n}})} < \frac{5}{(\frac{50}{\sqrt{100}})}) \\ = & P (\frac{- 5}{5} < z < \frac{5}{5}) \\ = & P (- 1 < z < 1) \\ = & P (z < 1) - P (z < - 1) \end{array}$

The probability of z less than –1 can be obtained using the excel formula “=NORM.S.DIST(–1,TRUE)”. The probability value is 0.1587.

The probability of z less than 1 can be obtained using the excel formula “=NORM.S.DIST(1,TRUE)”. The probability value is 0.8413.

The required probability value is,

$\begin{array}{rcl} P (- 5 < \bar{x} - μ < 5) & = & P (z < 1) - P (z < - 1) \\ = & 0.8413 - 0.1587 \\ = & 0.6826 \end{array}$

Thus, the probability of existence of a random sample (size 100) weight within $\pm$ 5 lb. of population mean weight is 0.6826.

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a) Find the probability of existence of a random sample (size 100) weight within + 5 lb. of population mean weight. b) For the same sample size, find the probability of the weight within + 10 lb. of population mean weight.

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