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Suppose that you form random samples of 25 from this distribution. Let X be the random variable of averages. Let ΣX be the random variable of sums.

Give the distribution of X.

User Weedoze
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Final answer:

The distribution of the sample means (X-bar) for a normally distributed population with mean 60 and standard deviation 3, from samples of 25, will be normally distributed with mean 60 and standard deviation 0.6. The sample sums (ΣX) also approximate normal distribution with mean 1500 and standard deviation 15. Calculations for specific probabilities or percentiles would use the properties of a normal distribution.

Step-by-step explanation:

When forming random samples from a normally distributed population where X has a mean (μ) of 60 and a standard deviation (σ) of 3, and using these samples to create a distribution of sample means (X-bar), the Central Limit Theorem comes into play.

This theorem states that the sample means will be normally distributed as well, especially as the sample size increases. The mean of X-bar is the same as the mean of the original distribution (μ), and the standard deviation of X-bar, also known as the standard error, is σ/√n, where n is the sample size. Since the sample size given is 25, the distribution of X-bar will be N(60, 3/√25) which simplifies to N(60, 0.6).

The sampling distribution of sums (ΣX) is also approaching a normal distribution according to the Central Limit Theorem. It will have a mean of (n)(μ) and a standard deviation of (σ)(√n), so with n=25, ΣX ~ N(1500, 15).

To perform specific calculations such as finding the probability P(x < 60), the 30th percentile for the mean, or the probability P(56 < x < 62), one would use the properties of the normal distribution, applying the appropriate z-values and the standard normal distribution table or a software that can handle statistical calculations.

User DoLoveSky
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