Answer:
1
The probability is [tex]P(\= X < 33) = 0.8531[/tex]
2
The probability is [tex]P(\= X > 30) = 0.3520[/tex]
Step-by-step explanation:
From the question we are told that
The population mean is [tex]\mu = 28.29[/tex]
The standard deviation is [tex]\sigma = 33.493[/tex]
The sample size is [tex]n = 56[/tex]
Generally the standard error for the sample mean [tex](\= x )[/tex] is mathematically evaluated as
[tex]\sigma _{\=x} = \frac{\sigma}{\sqrt{n} }[/tex]
substituting values
[tex]\sigma _{\=x} = \frac{33.493}{\sqrt{56} }[/tex]
[tex]\sigma _{\=x} = 4.48[/tex]
Apply central limit theorem[CLT] we have that
[tex]P(\= X < 33) = [z < \frac{33 - \mu }{\sigma_{\= x}} ][/tex]
substituting values
[tex]P(\= X < 33) = [z < \frac{33 - 28.29 }{4.48} ][/tex]
[tex]P(\= X < 33) = [z < 1.05 ][/tex]
From the z-table we have that
[tex]P(\= X < 33) = 0.8531[/tex]
For the second question
Apply central limit theorem[CLT] we have that
[tex]P(\= X > 30 ) = [z > \frac{30 - \mu }{\sigma_{\= x}} ][/tex]
substituting values
[tex]P(\= X < 33) = [z > \frac{30 - 28.29 }{4.48} ][/tex]
From the z-table we have that
[tex]P(\= X < 30) = 0.6480[/tex]
Thus
[tex]P(\= X > 30) = 1- P(\= X < 30) = 1- 0.6480[/tex]
[tex]P(\= X > 30) = 0.3520[/tex]
