SOLUTION: A random sample of size 43 is to be selected from a population that has a mean šœ‡ = 55 and a standard deviation šœŽ of 15. (b) Find the mean of this sampling distribution. (G

Algebra ->  Probability-and-statistics -> SOLUTION: A random sample of size 43 is to be selected from a population that has a mean šœ‡ = 55 and a standard deviation šœŽ of 15. (b) Find the mean of this sampling distribution. (G      Log On


   

Question 1171028: A random sample of size 43 is to be selected from a population that has a mean šœ‡ = 55 and a standard deviation šœŽ of 15.
(b) Find the mean of this sampling distribution. (Give your answer correct to nearest whole number.)
(c) Find the standard error of this sampling distribution. (Give your answer correct to two decimal places.)
(d) What is the probability that this sample mean will be between 43 and 51? (Give your answer correct to four decimal places.)
(e) What is the probability that the sample mean will have a value greater than 51? (Give your answer correct to four decimal places.)
(f) What is the probability that the sample mean will be within 3 units of the mean? (Give your answer correct to four decimal places.)
Answer by CPhill(1987) About Me  (Show Source):
You can put this solution on YOUR website!
Let's solve this problem step-by-step.
**Given Information:**
* Population mean (Ī¼) = 55
* Population standard deviation (Ļƒ) = 15
* Sample size (n) = 43
**(b) Find the Mean of the Sampling Distribution**
The mean of the sampling distribution of the sample mean (Ī¼xĢ„) is equal to the population mean (Ī¼).
* Ī¼xĢ„ = Ī¼ = 55
Therefore, the mean of the sampling distribution is 55.
**(c) Find the Standard Error of the Sampling Distribution**
The standard error (ĻƒxĢ„) is the standard deviation of the sampling distribution of the sample mean. It's calculated as:
* ĻƒxĢ„ = Ļƒ / āˆšn
* ĻƒxĢ„ = 15 / āˆš43
* ĻƒxĢ„ ā‰ˆ 15 / 6.5574
* ĻƒxĢ„ ā‰ˆ 2.2875
Rounded to two decimal places, the standard error is 2.29.
**(d) Probability that the Sample Mean is Between 43 and 51**
We need to find P(43 < xĢ„ < 51). First, we need to convert the sample means to z-scores:
* zā‚ = (43 - 55) / 2.29 = -12 / 2.29 ā‰ˆ -5.24
* zā‚‚ = (51 - 55) / 2.29 = -4 / 2.29 ā‰ˆ -1.75
Now, we need to find P(-5.24 < Z < -1.75).
* P(Z < -1.75) ā‰ˆ 0.0401
* P(Z < -5.24) ā‰ˆ 0 (very close to 0)
Therefore, P(-5.24 < Z < -1.75) = P(Z < -1.75) - P(Z < -5.24) ā‰ˆ 0.0401 - 0 ā‰ˆ 0.0401
**(e) Probability that the Sample Mean is Greater Than 51**
We need to find P(xĢ„ > 51). We already calculated the z-score for 51: z = -1.75.
* P(Z > -1.75) = 1 - P(Z < -1.75)
* P(Z > -1.75) ā‰ˆ 1 - 0.0401 = 0.9599
**(f) Probability that the Sample Mean is Within 3 Units of the Mean**
We need to find P(55 - 3 < xĢ„ < 55 + 3), which is P(52 < xĢ„ < 58).
* zā‚ = (52 - 55) / 2.29 = -3 / 2.29 ā‰ˆ -1.31
* zā‚‚ = (58 - 55) / 2.29 = 3 / 2.29 ā‰ˆ 1.31
We need to find P(-1.31 < Z < 1.31).
* P(Z < 1.31) ā‰ˆ 0.9049
* P(Z < -1.31) ā‰ˆ 0.0951
Therefore, P(-1.31 < Z < 1.31) = P(Z < 1.31) - P(Z < -1.31) ā‰ˆ 0.9049 - 0.0951 = 0.8098
**Answers:**
(b) 55
(c) 2.29
(d) 0.0401
(e) 0.9599
(f) 0.8098