The three scores are 86, 74, and an unknown score x. The average (i.e. the sum divided by how many there are) is 80, so we have
Multiply both sides by 3 and you get
Subtract 160 from both sides to get
Note: you could solve this exercise very quickly if you notice that the two numbers, 86 and 74, are symmetrical with respect to 80, which is the average. So, the third number must be the average.
Answer:
(a) The standard error of the mean in this experiment is $0.052.
(b) The probability that the sample mean is between $3.78 and $3.86 is 0.5587.
(c) The probability that the difference between the sample mean and the population mean is less than $0.01 is 0.5754.
(d) The likelihood that the sample mean is greater than $3.92 is 0.9726.
Step-by-step explanation:
According to the Central Limit Theorem if we have an unknown population with mean <em>μ</em> and standard deviation <em>σ</em> and appropriately huge random samples (<em>n</em> > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.
Then, the mean of the distribution of sample mean is given by,
And the standard deviation of the distribution of sample mean is given by,
The information provided is:
As <em>n</em> = 40 > 30, the distribution of sample mean is .
(a)
The standard error is the standard deviation of the sampling distribution of sample mean.
Compute the standard deviation of the sampling distribution of sample mean as follows:
Thus, the standard error of the mean in this experiment is $0.052.
(b)
Compute the probability that the sample mean is between $3.78 and $3.86 as follows:
Thus, the probability that the sample mean is between $3.78 and $3.86 is 0.5587.
(c)
If the difference between the sample mean and the population mean is less than $0.01 then:
Compute the value of as follows:
Thus, the probability that the difference between the sample mean and the population mean is less than $0.01 is 0.5754.
(d)
Compute the probability that the sample mean is greater than $3.92 as follows:
Thus, the likelihood that the sample mean is greater than $3.92 is 0.9726.