Answer: There are 61 business students must be randomly selected to estimate the mean monthly earnings of business students at one college.
Step-by-step explanation:
Since we have given that
mean = $135
Standard deviation = $538
At 95% level of confidence, z = 1.96
So, we get that
[tex]\bar{x}=z\dfrac{\sigma}{\sqrt{n}}\\135=1.96\times \dfrac{538}{\sqrt{n}}\\\\\dfrac{135}{1.96}=\dfrac{538}{\sqrt{n}}\\\\68.87=\dfrac{538}{\sqrt{n}}\\\\\sqrt{n}=\dfrac{538}{68.87}\\\\\sqrt{n}=7.8\\\\n=7.8^2\approx 61[/tex]
Hence, there are 61 business students must be randomly selected to estimate the mean monthly earnings of business students at one college.
The number of business students must be randomly selected is 61 students.
standard deviation (σ) = 538, sample size (n), margin of error (E) = 135
Confidence (C) = 95% = 0.95
α = 1 - C = 1 - 0.95 = 0.05
α/2 = 0.025
The z score of α/2 is the same as the z score of 0.475 (0.5 - 0.025) which is equal to 1.96.
[tex]E=z_\frac{\alpha}{2} *\frac{\sigma}{\sqrt{n} } 135\\\\135=1.96*\frac{538}{\sqrt{n} }\\\\n=61[/tex]
The number of business students must be randomly selected is 61 students.
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