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In the EAI sampling problem, the population mean is $ 71,500 and the population standard deviation is $5000. For n = 30, text{there is a} 0.4908 text{

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asked 2021-02-08
In the EAI sampling problem, the population mean is $ 71,500 and the population standard deviation is $5000. For \(n = 30,\ \text{there is a}\ 0.4908\ \text{probability of obtaining a sample mean within}\ \pm\ $\ 600\) of the population mean.
a) What is the probability that \(\bar{x}\) is within $ 600 of the population mean is a sample of size 60 is used (to 4 decimals)?
b) Answer part (a) for a sample of size 120 (to 4 decimals).

Answers (1)

2021-02-09

a) The probability that \(\bar{x}\)is within $600 of the population mean if a sample of size 60 used is,
\(\displaystyle{P}{\left(-{600}<\overline{{x}}-\mu<{600}\right)}={P}{\left(\frac{{-{600}}}{{{\left(\frac{5.000}{\sqrt{{60}}}\right)}}}<\frac{{\overline{{x}}-\mu}}{{{\left(\frac{\sigma}{\sqrt{{n}}}\right)}}}<\frac{600}{{{\left(\frac{5.000}{\sqrt{{60}}}\right)}}}\right)}\)
\(\displaystyle={P}{\left(\frac{{-{600}}}{{645.4972}}<{z}<\frac{60}{{645.4972}}\right)}\)
\(\displaystyle={P}{\left(-{0.93}<{z}<{0.93}\right)}\)
\(\displaystyle={P}{\left({z}<{0.93}\right)}-{P}{\left({z}<-{0.93}\right)}\)
The probability of z less than \(–0.93\ \text{can be obtained using the excel formula}\ \displaystyle“={N}{O}{R}{M}.{S}.{D}{I}{S}{T}{\left(–{0.93},{T}{R}{U}{E}\right)}\text{"}.\) The probability value is 0.1762.
The probability of z less than \(0.93\ \text{can be obtained using the excel formula}\ \displaystyle“={N}{O}{R}{M}.{S}.{D}{I}{S}{T}{\left({0.93},{T}{R}{U}{E}\right)}\text{"}\). The probability value is 0.8238.
The required probability value is,
\(\displaystyle{P}{\left(-{600}<\overline{{x}}-\mu<{600}\right)}={P}{\left({z}<{0.93}\right)}-{P}{\left({z}<-{0.93}\right)}\)
\(= 0.9238\ -\ 0.1762\)
\(= 0.6476\)
Thus, the probability that x- is within $600 of the population mean if a sample of size 60 used is 0.6476.
b) The probability that x- is within $600 of the population mean if a sample of size 120 used is,
\(\displaystyle{P}{\left(-{600}<\overline{{x}}-\mu<{600}\right)}={P}{\left(\frac{{-{600}}}{{{\left(\frac{5.000}{\sqrt{{120}}}\right)}}}<\frac{{\overline{{x}}-\mu}}{{{\left(\frac{\sigma}{\sqrt{{n}}}\right)}}}<\frac{600}{{{\left(\frac{5.000}{\sqrt{{120}}}\right)}}}\right)}\)
\(\displaystyle={P}{\left(\frac{{-{600}}}{{456.4355}}<{z}<\frac{60}{{456.4355}}\right)}\)
\(\displaystyle={P}{\left(-{1.31}<{z}<{131}\right)}\)
\(\displaystyle={P}{\left({z}<{1.31}\right)}-{P}{\left({z}<-{1.31}\right)}\)
The probability of z less than \(-1.31\ \text{can be obtained using the excel formula}\ \displaystyle“={N}{O}{R}{M}.{S}.{D}{I}{S}{T}{\left(–{1.31},{T}{R}{U}{E}\right)}\text{"}.\) The probability value is 0.0951.
The probability of z less than \(1.31\ \text{can be obtained using the excel formula}\ \displaystyle“={N}{O}{R}{M}.{S}.{D}{I}{S}{T}{\left({1.31},{T}{R}{U}{E}\right)}\text{"}.\) The probability value is 0.9049.
The required probability value is,
\(\displaystyle{P}{\left(-{600}<\overline{{x}}-\mu<{600}\right)}={P}{\left({z}<{1.31}\right)}-{P}{\left({z}<-{1.31}\right)}\)
\(= 0.9049\ -\ 0.0951\)
\(= 0.8098\)
Thus, the probability that x- is within $600 of the population mean if a sample of size 120 used is 0.8098.

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