Respuesta :
Answer:
95.44% confidence interval for the population mean is [19.2 , 20.8].
Step-by-step explanation:
We are given that a random sample of 144 observations has a mean of 20, a median of 21, and a mode of 22. The population standard deviation is known to equal 4.8.
Also, assuming that the data follows normal distribution.
So, the pivotal quantity for 95.44% confidence interval for the population mean is given by;
P.Q. = [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\bar X[/tex] = sample mean = 20
[tex]\sigma[/tex] = population standard deviation = 4.8
n = sample of obs. = 144
[tex]\mu[/tex] = population mean
So, 95.44% confidence interval for the population mean, [tex]\mu[/tex] is ;
P(-1.9991 < N(0,1) < 1.9991) = 0.9544
P(-1.9991 < [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < 1.9991) = 0.9544
P( [tex]-1.9991 \times {\frac{4.8}{\sqrt{144} }[/tex] < [tex]{\bar X - \mu}[/tex] < [tex]1.9991 \times {\frac{4.8}{\sqrt{144} }[/tex] ) = 0.9544
P( [tex]\bar X - 1.9991 \times {\frac{\sigma}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < [tex]\bar X + 1.9991 \times {\frac{\sigma}{\sqrt{n} }[/tex] ) = 0.9544
95.44% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X - 1.9991 \times {\frac{\sigma}{\sqrt{n} }[/tex] , [tex]\bar X + 1.9991 \times {\frac{\sigma}{\sqrt{n} }[/tex] ]
= [ [tex]20 - 1.9991 \times {\frac{4.8}{\sqrt{144} }[/tex] , [tex]20 + 1.9991 \times {\frac{4.8}{\sqrt{144} }[/tex] ]
= [19.2 , 20.8]
Therefore, 95.44% confidence interval for the population mean is [19.2 , 20.8].
