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Answer:
[tex] z= \frac{47 -50}{\frac{12}{\sqrt{36}}}=-1.5[/tex]
[tex] z= \frac{53 -50}{\frac{12}{\sqrt{36}}}=1.5[/tex]
And using a calculator, excel ir the normal standard table we have that:
[tex]P(47 \leq \bar X \leq 53) =P(-1.5 \leq Z \leq 1.5)[/tex]
And we can calculate the probability like this:[tex] P(-1.5 \leq Z \leq 1.5) = P(z<1.5) -P(Z<-1.5) =0.933-0.0668= 0.866[/tex]Step-by-step explanation:
A random sample of 36 observations has been drawn from a normal distribution with mean 50 and standard deviation 12. Find the probability that the sample mean is in the interval 47<=X<53. Is the assumption of normality important. Why?
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(50,12)[/tex]
Where [tex]\mu=50[/tex] and [tex]\sigma=12[/tex]
Since the distribution for X is normal then we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
We can find the probability required like this:
[tex] z= \frac{47 -50}{\frac{12}{\sqrt{36}}}=-1.5[/tex]
[tex] z= \frac{53 -50}{\frac{12}{\sqrt{36}}}=1.5[/tex]
And using a calculator, excel ir the normal standard table we have that:
[tex]P(47 \leq \bar X \leq 53) =P(-1.5 \leq Z \leq 1.5)[/tex]
And we can calculate the probability like this:
[tex] P(-1.5 \leq Z \leq 1.5) = P(z<1.5) -P(Z<-1.5) =0.933-0.0668= 0.866[/tex]
86.64% of the sample mean is in the interval would be between 47 and 53
The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:
[tex]z=\frac{x-\mu}{\sigma/\sqrt{n} } \\\\where\ \mu=mean,x=raw\ score,\sigma=standard\ deviation,n=sample\ size[/tex]
Given that μ = 50, σ = 12, n = 36:
For x = 47:
[tex]z=\frac{47-50}{12/\sqrt{36} } =-1.5\\\\\\For\ x=53:\\\\z=\frac{53-50}{12/\sqrt{36} } =1.5[/tex]
Therefore, From the normal distribution table: P(47 < x < 53) = P(-1.5 < z < 1.5) = P(z < 1.5) - P(z < -1.5) = 0.9332 - 0.0668 = 86.64%
Hence 86.64% of the sample mean is in the interval would be between 47 and 53
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