Answer:
Option e - 0.5597
Step-by-step explanation:
Given : Suppose that a random sample of size 49 is to be selected from a population with mean 49 and standard deviation 6.
To find : What is the approximate probability that will be more than 0.5 away from the population mean?
Solution :
Applying z-score formula,
[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
Here, [tex]\mu=49,\sigma=6, n=49[/tex]
As, probability that will be more than 0.5 away from the population mean
i.e. x=49.5,
[tex]z=\frac{49.5-49}{\frac{6}{\sqrt{49}}}[/tex]
[tex]z=\frac{0.5}{0.857}[/tex]
[tex]z=0.58[/tex]
In z-score table the value of z at 0.58 is 0.7190.
x=48.5,
[tex]z=\frac{48.5-49}{\frac{6}{\sqrt{49}}}[/tex]
[tex]z=\frac{-0.5}{0.857}[/tex]
[tex]z=-0.58[/tex]
In z-score table the value of z at -0.58 is 0.2810.
Now, probability that will be more than 0.5 away from the population mean is given by,
[tex]P=P(X>49.5)+P(X<48.5)[/tex]
[tex]P=1-P(X<49.5)+P(X<48.5)[/tex]
[tex]P=1-0.7190+0.2810[/tex]
[tex]P=0.562[/tex]
Therefore, option e is correct.
