Answer:
The standard error of the mean is 4.5.
Step-by-step explanation:
As we don't know the standard deviation of the population, we can estimate the standard error of the mean from the standard deviation of the sample as:
[tex]\sigma_{\bar{x}}\approx\frac{s}{\sqrt{n}}[/tex]
The sample is [30mins, 40 mins, 60 mins, 80 mins, 20 mins, 85 mins]. The size of the sample is n=6.
The mean of the sample is:
[tex]\bar{x}=\frac{1}n} \sum x_i =\frac{30+40+60+80+20+85}{6}=52.5[/tex]
The standard deviation of the sample is calculated as:
[tex]s=\sqrt{\frac{1}{n-1}\sum (x_i-\bar x)^2} \\\\ s=\sqrt{\frac{1}{5}\cdot ((30-52.5)^2+(40-52.5)^2+(60-52.5)^2+(80-52.5)^2+(20-52.5)^2+(85-52.5)^2}\\\\s=\sqrt{\frac{1}{5} *3587.5}=\sqrt{717.5}=26.8[/tex]
Then, we can calculate the standard error of the mean as:
[tex]\sigma_{\bar{x}}\approx\frac{s}{\sqrt{n}}=\frac{26.8}{6}= 4.5[/tex]
