Explain the relationship between the sampling distribution and sample size. 2. Explain the meaning of sampling error. 3. From a random sample of 40 divers, what is the probability that the mean time they spend in water is between 60 minutes and 90 minutes if the standard deviation is 2 minutes. 4. Explain the central limit theorem and state its importance in statistics. 5. A population consists of fire 4 numbers 25, 26, 30 and 22. a) Draw out all the possible samples of size 2 that can be obtained from this population. b) Determine the mean of the sampling distribution of means. c) Calculate the standard deviation of sampling distribution of means

Question 1

Answer · 2026-03-19T08:34:48.443Z

Question 1: The sampling distribution of the sample mean becomes narrower as the sample size increases because its variance is ^2 / n, which decreases as n grows. Larger samples lead to means closer to the population mean . For large n, the Central Limit Theorem ensures the sampling distribution is approximately normal. Question 2: A random sample is a subset where each population element has an equal chance of selection. Sampling error is the random difference between a sample statistic (e.g., x) and the population parameter (e.g., ). Its expected value is zero, and it shrinks with larger n. (The probability query on mean time within 2 minutes for times between 60-90 minutes lacks full details like distribution or n; assuming uniform or normal, use Chebyshev or z-scores accordingly, but cannot compute precisely.) Question 3: The Central Limit Theorem (CLT) states that for large n, the sampling distribution of x is approximately normal with mean and variance ^2/n, regardless of population shape. Importance: Enables inference (confidence intervals, hypothesis tests) using normal tables even for non-normal populations; crucial in statistics for n ≥ 30. Question 4: The population consists of four numbers: 22, 25, 26, 30. Question 5a: All possible samples of size n=2 (simple random sampling without replacement, N=4, so C(4,2)=6 equally likely samples): &\22,25\, &&\22,26\, &&\22,30\, \\ &\25,26\, &&\25,30\, &&\26,30\. Step 1: Compute sample means x for each. x_1 &= (22+25)/(2) = (47)/(2), \\ x_2 &= (22+26)/(2) = 24, \\ x_3 &= (22+30)/(2) = 26, \\ x_4 &= (25+26)/(2) = (51)/(2), \\ x_5 &= (25+30)/(2) = (55)/(2), \\ x_6 &= (26+30)/(2) = 28. Question 5b: Mean of the sampling distribution _x. Step 2: = (22+25+26+30)/(4) = (103)/(4). Sum of sample means: (47+48+52+51+55+56)/(2) = (309)/(2). _x = (309/2)/(6) = (309)/(12) = (103)/(4). The mean of the sampling distribution is (103)/(4). Question 5c: Standard deviation of the sampling distribution _x. Step 3: Population variance ^2. &= (103)/(4), \\ (22 - (103)/(4))^2 &= (-(15)/(4))^2 = (225)/(16), \\ (25 - (103)/(4))^2 &= (-(3)/(4))^2 = (9)/(16), \\ (26 - (103)/(4))^2 &= ((1)/(4))^2 = (1)/(16), \\ (30 - (103)/(4))^2 &= ((17)/(4))^2 = (289)/(16). Sum = (524)/(16), so ^2 = (524)/(16 × 4) = (131)/(16). Step 4: Use formula for without replacement: _x^2 = (^2)/(n) · (N-n)/(N-1) = (131/16)/(2) · (4-2)/(4-1) = (131)/(32) · (2)/(3) = (131)/(48). _x = sqrt((131)/(48)). The standard deviation of the sampling distribution is sqrt((131)/(48)). sqrt((131)/(48))

Question 2

Question 1: The sampling distribution of the sample mean becomes narrower as the sample size increases because its variance is ^2 / n, which decreases as n grows. Larger samples lead to means closer to the population mean . For large n, the Central Limit Theorem ensures the sampling distribution is approximately normal. Question 2: A random sample is a subset where each population element has an equal chance of selection. Sampling error is the random difference between a sample statistic (e.g., x) and the population parameter (e.g., ). Its expected value is zero, and it shrinks with larger n. (The probability query on mean time within 2 minutes for times between 60-90 minutes lacks full details like distribution or n; assuming uniform or normal, use Chebyshev or z-scores accordingly, but cannot compute precisely.) Question 3: The Central Limit Theorem (CLT) states that for large n, the sampling distribution of x is approximately normal with mean and variance ^2/n, regardless of population shape. Importance: Enables inference (confidence intervals, hypothesis tests) using normal tables even for non-normal populations; crucial in statistics for n ≥ 30. Question 4: The population consists of four numbers: 22, 25, 26, 30. Question 5a: All possible samples of size n=2 (simple random sampling without replacement, N=4, so C(4,2)=6 equally likely samples): &\22,25\, &&\22,26\, &&\22,30\, \\ &\25,26\, &&\25,30\, &&\26,30\. Step 1: Compute sample means x for each. x_1 &= (22+25)/(2) = (47)/(2), \\ x_2 &= (22+26)/(2) = 24, \\ x_3 &= (22+30)/(2) = 26, \\ x_4 &= (25+26)/(2) = (51)/(2), \\ x_5 &= (25+30)/(2) = (55)/(2), \\ x_6 &= (26+30)/(2) = 28. Question 5b: Mean of the sampling distribution _x. Step 2: = (22+25+26+30)/(4) = (103)/(4). Sum of sample means: (47+48+52+51+55+56)/(2) = (309)/(2). _x = (309/2)/(6) = (309)/(12) = (103)/(4). The mean of the sampling distribution is (103)/(4). Question 5c: Standard deviation of the sampling distribution _x. Step 3: Population variance ^2. &= (103)/(4), \\ (22 - (103)/(4))^2 &= (-(15)/(4))^2 = (225)/(16), \\ (25 - (103)/(4))^2 &= (-(3)/(4))^2 = (9)/(16), \\ (26 - (103)/(4))^2 &= ((1)/(4))^2 = (1)/(16), \\ (30 - (103)/(4))^2 &= ((17)/(4))^2 = (289)/(16). Sum = (524)/(16), so ^2 = (524)/(16 × 4) = (131)/(16). Step 4: Use formula for without replacement: _x^2 = (^2)/(n) · (N-n)/(N-1) = (131/16)/(2) · (4-2)/(4-1) = (131)/(32) · (2)/(3) = (131)/(48). _x = sqrt((131)/(48)). The standard deviation of the sampling distribution is sqrt((131)/(48)). sqrt((131)/(48))

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