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Academic Services 0 YouTube n Facebook-log In or AppsTop MINDTAP mplete: Chapter 7 Problem Set Due on Oct 13 at 5 PM EDT Attempts: Average: /6 2. Generating the sampling distribution of M Let's examine the mean of the numbers 1, 2, 3, 4,5,6,7, 8, 9, and 10 by drawing samples from these values, calaulating the mean of each sample, and then considering the sampling distribution of the mean. To do this, suppose you perform an experiment in which you roll a ten-sided die two times (or equivalently, rol two ten-sided dice one time) and calculate the mean of your sample. Remember that your population is the numbers 1, 2, 3, 4, 5 6, 7, 8, 9, and 10. The true mean (u) of the numbers 1,2, 3, 4, 5, 6, 7,8, 9, and 10 is 5.5 and the true standard deviation (o) is 33.61 5.80 2.87 r of possible different samples (each of size n-2) is the number of possibilities on the first roll (10) times of possibilities on the second roll (also 10), or 10(10)-100. If you collected all of these possible samples, the mean of your sampling distribution of means (HM) would equal 5.5and the standard deviation of your sampling distribution of means (that is, the standard error or ) would be The following chart shows the sampling distribution of the mean (M) for your experiment. Suppose you do this experiment once (that is, you roll the die two times). Use the chart to determine the probability that the mean df your two rolls is equal to the true mean, or P(M = ), is less than or equal to 1.5, or P(M s 1.5), is The probability that the mean of your two rolls is Type here to search O/10 e-@Explanation / Answer
Solution:
Here, we have to find out true population standard deviation for the given numbers.
Population SD = sqrt[(X - µ)/n]
We are given, µ = 5.5 and n = 10
Calculation table is given as below:
X
(X - µ)
(X - µ)2
1
-4.5
20.25
2
-3.5
12.25
3
-2.5
6.25
4
-1.5
2.25
5
-0.5
0.25
6
0.5
0.25
7
1.5
2.25
8
2.5
6.25
9
3.5
12.25
10
4.5
20.25
X = 55
(X - µ)2 = 82.5
µ = X / n = 55/10 = 5.5
2 = (X - µ)2 / n = 8.25
= sqrt(8.25) = 2.872281323
True standard deviation = = 2.87
Now, we have to find standard error which is given as below:
We are given n = 2
Standard error = SE = /sqrt(n) = 2.87/sqrt(2) = 2.029396
Now, we have to find P(M=µ) = P(M = 5.5)
By using normal approximation and continuity correction, we have to find
P(5<M<6) = P(M<6) – P(M<5)
For M=6
Z = (X - µ) / SE
Z = (6 – 5.5)/ 2.029396 = 0.246379
P(Z<0.246379) = P(M<6) = 0.597305 (By using z-table)
For M=5
Z = (5 – 5.5)/ 2.029396 = -0.24638
P(Z<-0.24638) = P(M<5) = 0.402695 (By using z-table)
P(5<M<6) = P(M<6) – P(M<5)
P(5<M<6) = 0.597305 - 0.402695
P(5<M<6) = 0.19461
Required probability = 0.19461
Now, we have to find P(M1.5)
Z = (1.5 – 5.5) / 2.029396
Z = -1.971029355
P(Z<-1.971029355) = P(M1.5) = 0.024360259
Required probability = 0.024360259
(By using z-table)
X
(X - µ)
(X - µ)2
1
-4.5
20.25
2
-3.5
12.25
3
-2.5
6.25
4
-1.5
2.25
5
-0.5
0.25
6
0.5
0.25
7
1.5
2.25
8
2.5
6.25
9
3.5
12.25
10
4.5
20.25
X = 55
(X - µ)2 = 82.5
µ = X / n = 55/10 = 5.5
2 = (X - µ)2 / n = 8.25
= sqrt(8.25) = 2.872281323