Question #310640

Il Directions: Perform what is being asked Solve the problem by showing the step-by-step process.


Situation: Miko, a DOH COVID-19 tracker, found out that the number' fected persons of COVID-19 in seven


low nisk provinces of the country are 1,9, 2, 12, 8, 7, and 1 Supposed to provinces are drawn as sample,


1. What is the mean (), variance (o) and standard deation of the population (c)?


2 How many different samples of size n =6 can be drawn from the pop station?


List them with their corresponding mean


Construct the sampling distribution of the sample menn

1
Expert's answer
2022-03-14T17:10:08-0400

Solution

1.

For a population;

mean μ=XiN\mu=\dfrac{\sum X_i}{N}


Variance σ2=(Xiμ)2N\sigma^2 =\dfrac{\sum (X_i-\mu)^2}{N}


Standard deviation σ=(Xiμ)2N\sigma=\sqrt\dfrac{\sum(X_i-\mu)^2}{N}


mean μ=1+9+2+12+8+7+17\mu = \dfrac{1+9+2+12+8+7+1}{7}


μ=5.71\mu=5.71


Variance


XiXiμ(Xiμ)214.7122.1814.7122.1823.7113.7671.291.6682.295.2493.2910.82126.2939.56115.40\def\arraystretch{1.5}\begin{array}{c:c:c} X_i&X_i -\mu&(X_i-\mu)^2\\\hline1&-4.71&22.18\\\hdashline1&-4.71&22.18\\\hdashline2&-3.71&13.76\\\hdashline7&1.29&1.66\\\hdashline8&2.29&5.24\\\hdashline9&3.29&10.82\\\hdashline12&6.29&39.56\\\hline \sum&&115.40\\\hline\end{array}


σ2=115.47=16.49\sigma^2=\dfrac{115.4}{7}=16.49


Standard deviation

σ=115.47=4.06\sigma=\sqrt\dfrac{115.4}{7} =4.06


2.


Population size N=7N=7

Sample size n=6n =6

Possible number of samples

=C76=(76)=7=C_7^6= \binom{7}{6}=7 Samples


no.SampleSamplemean11,1,2,7,8,94.6721,1,2,7,8,125.1731,1,2,7,9,125.3341,1,2,8,9,125.5051,1,7,8,9,126.3361,2,7,8,9,126.5071,2,7,8,9,126.50\def\arraystretch{1.5}\begin{array}{c:c:c}no.& Sample & Sample mean\\\hline1&1,1,2,7,8,9&4.67\\\hdashline2&1,1,2,7,8,12&5.17\\\hdashline3&1,1,2,7,9,12&5.33\\\hdashline4&1,1,2,8,9,12&5.50\\\hdashline5&1,1,7,8,9,12&6.33\\\hdashline6&1,2,7,8,9,12&6.50\\\hdashline7&1,2,7,8,9,12&6.50\\\hline\end{array}



Sampling distribution of the sample means


MeanProbability4.67175.17175.33175.50176.33176.5027\def\arraystretch{1.5}\begin{array}{c:c} Mean& Probability \\\hline4.67&\frac17\\\hdashline5.17&\frac17\\\hdashline5.33&\frac17\\\hdashline5.50&\frac17\\\hdashline6.33&\frac17\\\hdashline6.50&\frac27\\\hline\end{array}


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