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The credit card debts of all college students have a distribution that is skewed to the right with a mean of \(\$ 2840\) and a standard deviation of \(\$ 672\). Find the probability that the mean credit card debt for ? random sample of 36 college students would be a. between \(\$ 2600\) and \(\$ 2950\) b. less than \(\$ 3060\)

Short Answer

Expert verified
For part a, the probability that the mean credit card debt for a random sample of 36 college students is between \$2600 and \$2950 is given by the sum of the probabilities for the z-scores -2.14 and 0.98. For part b, the probability that the mean credit card debt is less than \$3060 is given by the area under the curve to the left of the z-score 1.96.

Step by step solution

01

Identify the given values

The problem gives the mean (\( µ = \$2840\)), the standard deviation (\( σ = \$672\)), and the sample size (\( n = 36\)). The sample mean \(\bar{x}\) is what we are finding the probability for. For part a, there are two values for \(\bar{x}\) to consider (\(\bar{x1} = \$2600\) and \(\bar{x2} = \$2950\)), and for part b, there is one (\(\bar{x3} = \$3060\)).
02

Find the standard error

The standard error (SE) is the standard deviation of the distribution of sample means. This is given by the formula \( SE = σ/√n = \$672/√36 = \$112\).
03

Calculate the z-scores

The z-score is calculated as follows: \( z = (\bar{x} - µ)/SE \). For part a, \( z1 = (\$2600 - \$2840)/\$112 = -2.14 \) and \( z2 = (\$2950 - \$2840)/\$112 = 0.98 \). For part b, \( z3 = (\$3060 - \$2840)/\$112 = 1.96 \).
04

Find the probabilities

Using the standard normal distribution table (Z-table) or a calculator with a built-in normal distribution function, find the probabilities corresponding to the z-scores. For part a, the probability \( P(\$2600 < \bar{x} < \$2950) = P(-2.14 < z < 0.98) \), which corresponds to the area under the curve between those two z-scores. This is done by finding the area from \( z = -2.14 \) to \( z = 0 \) and from \( z = 0 \) to \( z = 0.98 \), and then add these two probabilities together. For part b, the probability \( P(\bar{x} < \$3060) = P(z < 1.96) \), which corresponds to the area under the curve from \( -∞ \) to \( z = 1.96 \).
05

Evaluate the probabilities

Now it's time to interpret the results obtained, understanding what the probabilities mean in the real context of the problem. Here, they give the chances that the mean debt of a sample of 36 college students fall within certain value ranges.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

standard error
The standard error is a vital concept in statistics that helps us understand the variability of sample averages. It's particularly useful when we want to infer about a population based on a sample.

The standard error (SE) of the mean is calculated by dividing the population standard deviation (σ) by the square root of the sample size (n). So, in the formula: \[ SE = \frac{σ}{\sqrt{n}} \] you can see that as the sample size increases, the standard error decreases.

This makes theoretical sense since larger samples provide more reliable estimates of the population mean. For our exercise, with a standard deviation of 672 and a sample size of 36, the SE comes out to be 112. This value tells us how much the sample mean is expected to fluctuate above or below the population mean.
z-score
The z-score is a measure of how many standard deviations an element is from the mean. A z-score can tell us whether a particular data point is below or above the average.

In mathematical terms, for a given sample mean \(\bar{x}\), the z-score is calculated using:\[ z = \frac{(\bar{x} - µ)}{SE} \] Here, \(µ\) is the population mean, and \(SE\) is the standard error.

In our exercise example, by substituting these values, we calculated z-scores for different sample means to understand their relative positioning to the mean credit card debt.

This helped us find out how probable it is for the sample mean to fall within given ranges. For instance, a negative z-score means a sample mean is less than the average, while a positive z-score means it's greater.
normal distribution
A normal distribution is a fundamental concept in statistics that depicts how data is spread. It is symmetrical and bell-shaped, centering around the mean. Importantly, it allows for easy calculation of probabilities associated with various data points.

In our context, while the original data is skewed, the Central Limit Theorem tells us that the distribution of the sample mean will tend to be normal as the sample size increases. This is crucial because it allows us to use the normal distribution to approximate and calculate probabilities.

Most z-tables or statistical software rely on this property to allow us to find probabilities associated with given z-scores. The exercise required us to understand how sample means distribute themselves normally, even when the population data doesn't.
probability calculation
Probability calculation using the normal distribution is a powerful tool in statistics. It lets us determine the likelihood of occurrence of an event. Once we have z-scores for our sample means, we can use a standard normal distribution table, or statistical software, to find corresponding probabilities.

In the step-by-step solution, we calculated the probabilities that a sample mean would fall between \(\(2600\) and \(\\)2950\), or be less than \(\$3060\).

By looking up each z-score in a Z-table:
  • The table tells the combined area (probability) for z-scores smaller than that value.
  • For a range, we find probabilities for each endpoint and subtract them to get the probability in-between.
Understanding these probabilities helps in gauging how credit card debts of college students might realistically disperse across samples.

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