The Office of Student Services at a large western state university maintains information on the study habits of its full-time students. Their studies indicate that the mean amount of time undergraduate students study per week is 20 hours. The hours studied follows the normal distribution with a population standard deviation of six hours. Suppose we select a random sample of 144 current students. What is the standard error of the mean

There is a probability of 77% that the mean of this sample is between 19.25 hours and 21.0 hoursZ 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\ x=raw\ score,\mu=mean,\sigma=standard\ deviation,n=sample\ size[/tex]Given that μ = 20, σ = 7, n = 125.For x = 19.25:[tex]z=\frac{19.25-20}{7/\sqrt{125} } =-1.20\\\\\\For\ x=21:\\\\z=\frac{21-20}{7/\sqrt{125} } =0.62[/tex]From the normal distribution table, P(-1.20 < z < 0.62) = P(z < 0.62) - P(z < -1.2) = 0.8849 - 0.1151 = 0.7698 = 77%There is a probability of 77% that the mean of this sample is between 19.25 hours and 21.0 hoursFind out more on z score at: