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Answer :
Using the central limit theorem, we approximate the distribution of the sample mean and find the probability that it is at most 67kg.
To solve this problem, we need to find the probability that the sample mean of 9 students' weights is at most 67kg. Since the weights follow a normal distribution with a mean of 65kg and a variance of 4kg, we can use the central limit theorem to approximate the distribution of the sample mean. The sample mean is also normally distributed with the same mean of 65kg and a standard deviation of 4kg/sqrt(9) = 4/3kg.
To find the probability that the sample mean is at most 67kg, we need to find the area under the normal distribution curve to the left of 67kg. To do this, we can standardize the value of 67kg using z-score formula, which gives us (67-65)/(4/3) = 2.25. We can then find the corresponding area under the standard normal distribution curve using a z-table or a calculator. The probability is approximately 0.9872 or 98.72%, which is closest to option d. 0.9772.
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The probability that the sample mean of the 9 randomly chosen students is at most 67kg is 0.9987. So, The correct answer is b. 0.9987.
Now for the probability that the sample mean of 9 randomly chosen students from U.O.P. is at most 67kg, use the Central Limit Theorem.
Given that the population mean is 65kg and the variance is 4kg.
Hence the standard deviation, which is the square root of the variance:
√4 = 2kg.
Next, we calculate the Standard Error of the Mean (SEM) by dividing the standard deviation by the square root of the sample size:
2kg / √9
= 2kg / 3
= 0.67kg.
Now, use the Z-table to find the probability that the sample mean is at most 67kg.
We calculate the Z-score using the formula:
Z = (sample mean - population mean) / SEM.
Z = (67kg - 65kg) / 0.67kg
z ≈ 2.985
Using the Z-table or a calculator, the probability corresponding to a Z-score of 2.985 is 0.9987.
Therefore, the probability that the sample mean of the 9 randomly chosen students is at most 67kg is 0.9987.
So, The correct answer is b. 0.9987.
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