91Ó°ÊÓ

The upper leg length of 20 - to 29 -year-old males is normally distributed with a mean length of \(43.7 \mathrm{~cm}\) and a standard deviation of \(4.2 \mathrm{~cm} .\) Source: "Anthropometric Reference Data for Children and Adults: U.S. Population, 1999-2002"; Volume 361, July 7, 2005. (a) What is the probability that a randomly selected 20 - to 29 -yearold male has an upper leg length that is less than \(40 \mathrm{~cm} ?\) (b) A random sample of 9 males who are 20 to 29 years old is obtained. What is the probability that the mean upper leg length is less than \(40 \mathrm{~cm} ?\) (c) What is the probability that a random sample of 12 males who are \(20-29\) years old results in a mean upper leg length that is less than \(40 \mathrm{~cm} ?\) (d) What effect does increasing the sample size have on the probability? Provide an explanation for this result. (e) A random sample of 15 males who are \(20-29\) years old results in a mean upper leg length of \(46 \mathrm{~cm} .\) Do you find this result unusual? Why?

Step by step solution

01

Given Data

The mean upper leg length is \(\bar{x} = 43.7 \text{ cm}\) and the standard deviation is \(\text{SD} = 4.2 \text{ cm}\).

02

Calculate the Z-Score for Part (a)

To find the probability that a randomly selected male has an upper leg length less than 40 cm, compute the Z-score using the formula \(Z = \frac{X - \bar{x}}{\text{SD}}\). Plug in the values: \(Z = \frac{40 - 43.7}{4.2} = -0.88\).

03

Find Probability for Part (a)

Use a Z-table or calculator to find the probability corresponding to \(Z = -0.88\) which is approximately \(\text{P}(Z < -0.88) \approx 0.1894\).

04

Calculate the Standard Error for Part (b)

Determine the standard error for the mean of a sample of 9 males using \( \text{SE} = \frac{\text{SD}}{\root n} = \frac{4.2}{\root 9} = 1.4 \text{ cm}\).

05

Calculate the Z-Score for Part (b)

For the mean length less than 40 cm, use \(Z = \frac{40 - 43.7}{1.4} = -2.64\).

06

Find Probability for Part (b)

Using a Z-table, the probability corresponding to \(Z = -2.64\) is approximately \(\text{P}(Z < -2.64) \approx 0.0041\).

07

Calculate the Standard Error for Part (c)

Determine the standard error for the mean of a sample of 12 males using \( \text{SE} = \frac{\text{SD}}{\root n} = \frac{4.2}{\root 12} = 1.21 \text{ cm}\).

08

Calculate the Z-Score for Part (c)

For the mean length less than 40 cm, use \(Z = \frac{40 - 43.7}{1.21} = -3.06\).

09

Find Probability for Part (c)

Using a Z-table, the probability corresponding to \(Z = -3.06\) is approximately \(\text{P}(Z < -3.06) \approx 0.0011\).

10

Explain Effect of Increasing Sample Size for Part (d)

As the sample size increases, the standard error decreases. This results in a larger Z-score, thus decreasing the probability of obtaining a mean less than a certain value. Hence, increasing the sample size results in a lower probability of obtaining any extreme result.

11

Determine if Result is Unusual for Part (e)

Calculate the Z-score for mean 46 cm with sample size 15 using \( \text{SE} = \frac{\text{SD}}{\root n} = \frac{4.2}{\root 15} = 1.08 \text{ cm}\) and then \(Z = \frac{46 - 43.7}{1.08} = 2.13\). Probability for \(Z > 2.13\) is approximately \(\text{P}(Z > 2.13) \approx 0.0166\) which is low, indicating that the result is unusual.

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Key Concepts

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

Probability

Probability is a measure of how likely an event is to occur. It ranges from 0 (an impossible event) to 1 (a certain event). In the context of normal distribution problems, we often seek the probability of a specific outcome within a range of values. To find this, we use the Z-score, which helps standardize the data and connects it to the standard normal distribution. The probabilities are often derived from Z-tables or computational tools that provide the area under the curve for specific Z-scores.
For instance, in the given exercise, we want to find the probability that an upper leg length is less than 40 cm. We compute a Z-score and then find the corresponding probability from the Z-table.

Z-Score

The Z-score indicates how many standard deviations an element is from the mean. It is calculated using the formula: \( Z = \frac{X - \bar{x}}{\text{SD}} \), where \( X \) is the value, \( \bar{x} \) is the mean, and \( \text{SD} \) is the standard deviation.
In our example, the Z-score for an upper leg length of 40 cm is calculated as \( Z = \frac{40 - 43.7}{4.2} = -0.88 \). This tells us that 40 cm is -0.88 standard deviations below the mean. Z-scores allow us to use standard normal distribution tables to find probabilities associated with a value.

Standard Error

Standard error measures the spread of the sample mean estimates around the population mean. It decreases as the sample size increases, indicating better precision in our estimates.
The standard error is given by the formula: \( \text{SE} = \frac{\text{SD}}{\root n} \), where \( \text{SD} \) is the population standard deviation and \( n \) is the sample size.
For a sample of 9 males, the standard error is calculated as \( \frac{4.2}{\root 9} = 1.4 \) cm. For a sample of 12 males, it is \( \frac{4.2}{\root 12} = 1.21 \) cm. A smaller standard error leads to larger Z-scores for the same sample means, affecting the calculated probabilities.

Sample Size Effect

The sample size significantly influences the standard error and, consequently, the probability of observing certain sample mean values. As sample size increases, the standard error decreases, meaning our estimate of the mean becomes more precise.
This results in larger Z-scores for the same differences between sample mean and population mean, thus leading to lower probabilities of extreme sample means. This explains why, as shown in the exercise, the probability of getting an upper leg length less than 40 cm decreases with larger sample sizes.

Unusual Results

Results can be deemed unusual if their probability is very low. Typically, results with a probability less than 0.05 (5%) are considered unusual.
In our exercise, a sample mean upper leg length of 46 cm for a sample size of 15 males yields a Z-score of 2.13. The corresponding probability is approximately 0.0166, or 1.66%, which is quite low. This makes the result unusual, indicating that such a sample mean is not typical under normal distribution with given parameters.