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Chapter 6: Problem 8

Using the Central Limit Theorem assume that females have pulse rates 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 "Body Data" 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) 0.2742b) 0.1003c) The original distribution is normal.

Step by step solution

01

Find Z-scores for Individual Female - Part (a)

To find the probability of an individual female's pulse rate being between 78 and 90 beats per minute, calculate the Z-scores using the formula: \[ Z = \frac{X - \mu}{\sigma} \] where X is the value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. For X = 78, \[ Z_1 = \frac{78 - 74.0}{12.5} = \frac{4}{12.5} = 0.32 \] For X = 90, \[ Z_2 = \frac{90 - 74.0}{12.5} = \frac{16}{12.5} = 1.28 \]
02

Use Z-table - Part (a)

Look up the Z-scores in the standard normal distribution table (Z-table). Z-score 0.32 corresponds to a cumulative probability of approximately 0.6255 and Z-score 1.28 corresponds to a cumulative probability of approximately 0.8997. The probability of a pulse rate being between 78 and 90 beats per minute is \( P(78 < X < 90) = P(Z_2) - P(Z_1) = 0.8997 - 0.6255 = 0.2742 \).
03

Find Z-scores for Sample Mean - Part (b)

To find the probability for the mean of 16 females, use the Central Limit Theorem. The standard deviation of the sample mean (\(\sigma_{\bar{X}}\)) is given by: \[ \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{12.5}{\sqrt{16}} = \frac{12.5}{4} = 3.125 \] Calculate the Z-scores for the sample means between 78 and 90: For X = 78, \[ Z_1 = \frac{78 - 74.0}{3.125} = \frac{4}{3.125} = 1.28 \] For X = 90, \[ Z_2 = \frac{90 - 74.0}{3.125} = \frac{16}{3.125} = 5.12 \]
04

Use Z-table - Part (b)

Look up the Z-scores in the standard normal distribution table. Z-score 1.28 corresponds to a cumulative probability of approximately 0.8997, and Z-score 5.12 is essentially 1 (since it is very large). The probability of the sample mean being between 78 and 90 beats per minute is \( P(78 < \bar{X} < 90) = P(Z_2) - P(Z_1) = 1 - 0.8997 = 0.1003. \)
05

Explain Use of Normal Distribution - Part (c)

The Central Limit Theorem states that the distribution of the sampling mean approaches a normal distribution, regardless of the shape of the original distribution, provided the sample size is sufficiently large (n 鈮� 30) or the original distribution is normal. Since the pulse rates are normally distributed, the sample mean of 16 females will also be normally distributed.

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Key Concepts

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

Normally Distributed
When we say something is 'normally distributed,' we refer to a specific type of continuous probability distribution. In a normal distribution, most values cluster around a central region, with the probabilities tapering off symmetrically towards both extremes. Imagine plotting a large amount of data on a graph: this data would create a bell-shaped curve, known as the Gaussian or bell curve. The highest point on the bell curve represents the mean, or average value. Key features of a normal distribution include being symmetric about the mean, having a single peak (unimodal), and having tails that approach the horizontal axis asymptotically.
Z-scores
A Z-score measures the number of standard deviations a data point is from the mean of a data set. Z-scores help to standardize data, allowing different data sets to be compared. The formula to calculate a Z-score is:

\[ Z = \frac{X - \text{{mean}}}{\text{{standard deviation}}} \]
For example, if the mean pulse rate is 74 beats per minute with a standard deviation of 12.5, and we want to calculate the Z-score for 78 beats per minute, it would be:

\[ Z = \frac{78 - 74}{12.5} = 0.32 \]
A Z-score of 0.32 indicates that 78 bpm (beats per minute) is 0.32 standard deviations above the mean.
Standard Normal Distribution
The standard normal distribution is a special case of the normal distribution. It has a mean of 0 and a standard deviation of 1. When any data from a normal distribution is standardized (converted to Z-scores), it follows a standard normal distribution. This makes it easier to calculate probabilities because standard normal distribution tables (Z-tables) are widely available.
For instance, if a female's pulse rate follows a normal distribution with a mean of 74 bpm and standard deviation of 12.5 bpm, converting any pulse rate to a Z-score allows us to use the Z-table to find probabilities.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. It's a crucial metric in statistics because it tells us how spread out the values are around the mean. A low standard deviation means that values tend to be close to the mean, whereas a high standard deviation means that values are spread out over a wide range. The formula for standard deviation for a sample is:

\[ s = \frac{\text{{sum of squared deviations}}}{n-1} \] where 'n' is the number of observations.
In our example, the standard deviation of pulse rates is 12.5 bpm. This indicates the typical distance of pulse rate values from the mean (74 bpm) is 12.5 bpm.
Sample Mean
The sample mean is the average of observations from a sample, which is a subset of a population. It's used to estimate the population mean. The formula for the sample mean is the sum of all sample values divided by the number of samples. When using the Central Limit Theorem (CLT), the mean of sample means will be equal to the population mean, and the standard error (standard deviation of sample means) is calculated using:

\[ \text{{Standard Error}} = \frac{\text{{Population Standard Deviation}}}{\text{{Square Root of Sample Size}}} \]
For instance, in part (b) of the problem, the standard error for 16 females' pulse rates is calculated as follows:
\[ \text{{SE}} = \frac{12.5}{\root 2{16}} = 3.125 \]
This implies the distribution of sample means will be tighter than the original distribution by a factor dependent on the sample size.

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Most popular questions from this chapter

Unbiased Estimators Data Set 4 "Births" in Appendix B includes birth weights of 400 babies. If we compute the values of sample statistics from that sample, which of the following statistics are unbiased estimators of the corresponding population parameters: sample mean; sample median; sample range; sample variance; sample standard deviation; sample proportion?

Loading Aircraft Before every flight, the pilot must verify that the total weight of the load is less than the maximum allowable load for the aircraft. The Bombardier Dash 8 aircraft can carry 37 passengers, and a flight has fuel and baggage that allows for a total passenger load of 6200 lb. The pilot sees that the plane is full and all passengers are men. The aircraft will be overloaded if the mean weight of the passengers is greater than \(6200 \mathrm{lb} / 37=167.6 \mathrm{lb}\). What is the probability that the aircraft is overloaded? Should the pilot take any action to correct for an overloaded aircraft? Assume that weights of men are normally distributed with a mean of 189 lb and a standard deviation of 39 lb (based on Data Set I "Body Data" in Appendix B).

Do the following: If the requirements of \(n p \geq 5\) and \(n q \geq 5\) are both satisfied, estimate the indicated probability by using the normal distribution as an approximation to the binomial distribution; if \(n p < 5\) or n \(q < 5,\) then state that the normal approximation should not be used. With \(n=8\) births and \(p=0.512\) for a boy, find \(P\) (exactly 5 boys).

Find the indicated area under the curve of the standard normal distribution; then convert it to a percentage and fill in the blank. The results form the basis for the range rule of thumb and the empirical rule introduced in Section 3-2. About _____ \(\%\) of the area is between \(z=-3\) and \(z=3\) (or within 3 standard deviations of the mean).

The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. a. In a letter to "Dear Abby," a wife claimed to have given birth 308 days after a brief visit from her husband, who was working in another country. Find the probability of a pregnancy lasting 308 days or longer. What does the result suggest? b. If we stipulate that a baby is premature if the duration of pregnancy is in the lowest \(3 \%\) find the duration that separates premature babies from those who are not premature. Premature babies often require special care, and this result could be helpful to hospital administrators in planning for that care.
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