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Assume that females have pulserates that are normally distributed with a mean of 74.0 beats per minute and a standard deviation of 12.5 beats per minute (based on Data Set 1 鈥淏ody Data鈥 in Appendix B). a. If 1 adult female is randomly selected, find the probability that her pulse rate is between 78 beats per minute and 90 beats per minute. b. If 16 adult females are randomly selected, find the probability that they have pulse rates with a mean between 78 beats per minute and 90 beats per minute. c. Why can the normal distribution be used in part (b), even though the sample size does not exceed \(30 ?\)

Short Answer

Expert verified
a: 0.2742, b: 0.1003, c: Normal distribution applies due to Central Limit Theorem as underlying pulse rate distribution is normal.

Step by step solution

01

Title - Identify given data for part (a)

Mean \(\boxed{\text{渭}}\) = 74 beats per minute, Standard Deviation \(\boxed{\text{蟽}}\) = 12.5 beats per minute. We need to find the probability that one randomly selected female has a pulse rate between 78 and 90 beats per minute.
02

Title - Convert raw scores to z-scores for part (a)

First, convert the pulse rates to z-scores using the formula \( z = \frac{x - 渭}{蟽} \).For 78 beats per minute: \(\boxed{\frac{78 - 74}{12.5} = 0.32} \).For 90 beats per minute: \(\boxed{\frac{90 - 74}{12.5} = 1.28} \).
03

Title - Use z-table to find probabilities for part (a)

Find the probability corresponding to the z-scores. \( P(0.32 \le Z \le 1.28) \). Using z-tables: \(P(Z \le 1.28) 鈮 0.8997 \). \(P(Z \le 0.32) 鈮 0.6255 \). Thus, \(P(0.32 \le Z \le 1.28) = 0.8997 - 0.6255 = 0.2742 \).
04

Title - Identify given data for part (b)

Mean \(\boxed{\text{渭}}\) = 74 beats per minute, Sample Size = 16, Standard Deviation \(\boxed{\text{蟽}}\) = 12.5 beats per minute. We need to find the probability that the mean pulse rate of 16 randomly selected females is between 78 and 90 beats per minute.
05

Title - Calculate standard error for part (b)

Standard Error (SE) is calculated as \( SE = \frac{蟽}{\text{鈭歯}} \), where \( \text{n} = 16 \). So, \( SE = \frac{12.5}{4} = 3.125 \).
06

Title - Convert mean pulse rates to z-scores for part (b)

For 78 beats per minute: \( z = \frac{78 - 74}{3.125} = 1.28 \). \ For 90 beats per minute: \( z = \frac{90 - 74}{3.125} = 5.12 \).
07

Title - Use z-table to find probabilities for part (b)

Find the probability corresponding to the z-scores. \( P(1.28 \le Z \le 5.12) \). Using z-tables: \(P(Z \le 5.12) 鈮 1 \). \(P(Z \le 1.28) 鈮 0.8997 \). Thus, \(P(1.28 \le Z \le 5.12) = 1 - 0.8997 = 0.1003 \).
08

Title - Explain the use of normal distribution in part (b)

The Central Limit Theorem states that the sampling distribution of the sample mean will be normally distributed if the sample size is sufficiently large (n >= 30) or if the underlying distribution is normal. Since the individual pulse rates are normally distributed, the sample mean distribution is also normally distributed, even with a sample size of 16.

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Key Concepts

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

z-scores
To find probabilities for normally distributed data, a common method is to convert the data points into z-scores. A z-score represents the number of standard deviations a data point is from the mean. You can calculate the z-score using the formula:

\( z = \frac{x - 渭}{蟽} \), where \( x \) is the data point, \( 渭 \) is the mean, and \( 蟽 \) is the standard deviation.

For example, if we want to find the probability that an adult female has a pulse rate between 78 and 90 beats per minute, we first convert these pulse rates into z-scores using the given mean of 74 and standard deviation of 12.5. So:

\( z = \frac{78 - 74}{12.5} = 0.32 \) and \( z = \frac{90 - 74}{12.5} = 1.28 \).

These z-scores help us to find the corresponding probabilities using a z-table.
Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental principle in statistics. It states that the sampling distribution of the sample mean will approximate a normal distribution, regardless of the shape of the population distribution, provided the sample size is sufficiently large (typically \( n \geq 30 \)).

Even for smaller sample sizes, like 16 in our exercise, if the underlying population distribution is normal, the CLT ensures that the distribution of the sample mean will be approximately normal.

This theorem is key to making inferences about population parameters. In our example, although our sample size of pulse rates is only 16, we can use the normal distribution due to the original data (pulse rates) being normally distributed.
probability calculation
Probability calculation in normal distribution involves finding the area under the curve between two z-scores. This area represents the likelihood of the data falling within a specified range.

For instance, to calculate the probability that a randomly selected adult female has a pulse rate between 78 and 90 beats per minute, we convert these values to z-scores and then use a z-table to find the probabilities.

The z-scores for 78 and 90 are 0.32 and 1.28, respectively. Using the z-table, we find that \( P(Z \leq 1.28)=0.8997 \) and \( P(Z \leq 0.32)=0.6255 \).

Therefore, \( P(0.32 \leq Z \leq 1.28) = 0.8997 - 0.6255 = 0.2742 \).

This means there's a 27.42% chance that a single adult female's pulse rate will fall between 78 and 90 beats per minute.
standard error
Standard Error (SE) measures the variability of sample means around the population mean. It provides insights into the precision of our sample mean as an estimate of the population mean.

The formula to calculate standard error is:

\( SE = \frac{蟽}{\text{鈭歯}} \), where \( 蟽 \) is the standard deviation of the population and \( n \) is the sample size.

In our exercise, with a standard deviation of 12.5 and a sample size of 16, the SE is: \( SE = \frac{12.5}{4} = 3.125 \).

This smaller SE indicates that the sample mean will be close to the population mean, enhancing our confidence in the sample mean as a reliable estimate.
sampling distribution
The sampling distribution of the sample mean is the distribution of means from all possible samples of a particular size taken from a population. It鈥檚 crucial when making inferences about the population mean.

According to the Central Limit Theorem, the sampling distribution tends to be normal if the sample size is large or if the population distribution is normal. In our scenario, even with a sample size of 16, the normally distributed pulse rates ensure that the sampling distribution of the mean is also normal.

This distribution helps us determine probabilities for sample means. For instance, in part (b) of our exercise, we calculated the z-scores for the sample mean pulse rates and found the probability using the z-table.

Because the sampling distribution is normal, we could confidently calculate the probability correctly, showcasing its importance in statistical analysis.

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