Professor Smith conducted a class exercise in which students ran a computer program to generate random samples from a population that had a mean of 50 and a standard deviation of 9mm. Each of Smith's students took a random sample of size n and calculated the sample mean. Smith found that about 68% of the students had sample means between 48.5 and 51.5 mm. What was n? ( assume that n is large enogh that the Central Limiot Theorem is applicable)

To find the value of n, we need to use the Central Limit Theorem and the properties of the normal distribution.

According to the Central Limit Theorem, regardless of the shape of the population distribution, the distribution of sample means will be approximately normally distributed for large enough sample sizes.

Given that about 68% of the students had sample means between 48.5 and 51.5 mm, we can infer that this range represents one standard deviation below and above the population mean. In a normal distribution, this range corresponds to approximately 68% of the data.

Since the standard deviation of the population is 9mm, we can set up the following equation:

51.5 - 50 = 9 / sqrt(n)

To simplify the equation, we square both sides:

(51.5 - 50)^2 = (9 / sqrt(n))^2

1.5^2 = (9^2) / n

2.25 = 81 / n

Cross multiplying and rearranging gives us:

2.25n = 81

n = 81 / 2.25

n ≈ 36

Therefore, the approximate value of n is 36.

To find the value of n, we need to use the Central Limit Theorem (CLT) and the properties of the normal distribution.

The Central Limit Theorem states that the distribution of sample means approaches a normal distribution with a mean equal to the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size (n). So, for our problem, we can write:

Standard deviation of sample means = Population standard deviation / sqrt(n)

Given that 68% of the students had sample means between 48.5 and 51.5 mm, it means that this interval corresponds to one standard deviation on both sides of the mean. In other words:

Mean ± 1 standard deviation

Since the mean is 50 mm, we can write:

48.5 mm ≤ Mean ≤ 51.5 mm

To find the standard deviation of the sample means, we can set up the following equation:

51.5 mm - 50 mm = (Population standard deviation) / sqrt(n)

Then, we can solve this equation for n:

sqrt(n) = (Population standard deviation) / (51.5 mm - 50 mm)

sqrt(n) = 9 mm / 1.5 mm

Taking the square root of both sides, we get:

sqrt(n) ≈ 6

Squaring both sides of the equation:

n ≈ 36

Hence, the value of n is approximately 36.

The standard deviation of n-sample means is sigma/sqrt(n). In your case sigma = 9 for the single-sample distribution and the 1.5 for the groups of n. Therefore

1.5 = 9/sqrt(n)
sqrt n = 9/1.5 = 6 and n = 36