Problem 80 - Applied Statistics and Probability for Engineers

Question

In 2002 ,the average height of a woman aged \(20-74\) years was 64 inches with an increase of approximately 1 inch from 1960 (http://usgovinfo.about.com/od/healthcare). Suppose the height of a woman is normally distributed with a standard deviation of two inches. (a) What is the probability that a randomly selected woman in this population is between 58 inches and 70 inches? (b) What are the quartiles of this distribution? (c) Determine the height that is symmetric about the mean that includes \(90 \%\) of this population. (d) What is the probability that five women selected at random from this population all exceed 68 inches?

Step-by-Step Solution

Answer

(a) 0.9974, (b) Q1: 62.652 inches, Q3: 65.348 inches, (c) 60.71 to 67.29 inches, (d) ≈ 6.67 × 10⁻⁸.

Step 1: Understanding Normal Distribution

We are given a normal distribution with a mean of 64 inches and a standard deviation of 2 inches. This means the distribution of women's heights in this population is modeled by the normal distribution: \(X \sim N(64, 2^2)\).

Step 1: Calculating the Probability for Part (a)

To find the probability that a randomly selected woman is between 58 and 70 inches, we need to calculate \(P(58 < X < 70)\). We do this using the standard normal distribution (Z-distribution). First, convert the heights to z-scores using the formula \(Z = \frac{{X - \mu}}{\sigma}\)- For 58 inches: \(Z_1 = \frac{{58 - 64}}{2} = -3\)- For 70 inches: \(Z_2 = \frac{{70 - 64}}{2} = 3\)Now, we find \(P(-3 < Z < 3)\) using standard normal distribution tables or software, which gives approximately 0.9974.

Step 2: Finding the Quartiles for Part (b)

The quartiles (Q1 and Q3) of a normal distribution are the 25th and 75th percentiles. We find the z-scores corresponding to these percentiles: - The z-score for Q1 (25th percentile) is approximately -0.674.- The z-score for Q3 (75th percentile) is approximately 0.674.Using the z-score formula \(X = Z \sigma + \mu\):- Q1: \(X_1 = -0.674 \times 2 + 64 = 62.652\) inches- Q3: \(X_3 = 0.674 \times 2 + 64 = 65.348\) inches

Step 3: Calculating Part (c) - Symmetric Height for 90% Population

We need the central range that includes 90% of the distribution. This means we want the middle 90%, or that 5% is left on both tails of the distribution. The corresponding z-scores for the tails (5th and 95th percentiles) are approximately -1.645 and 1.645, respectively. Thus, the symmetric height range is:- Lower Limit: \(X = -1.645 \times 2 + 64 = 60.71\) inches- Upper Limit: \(X = 1.645 \times 2 + 64 = 67.29\) inches

Step 4: Calculating Part (d) - Probability of Five Women Exceeding 68 Inches

First, find the probability that one woman exceeds 68 inches. Calculate the z-score for 68 inches: - \(Z = \frac{{68 - 64}}{2} = 2\)The probability of a woman being over 68 inches is \(P(Z > 2)\). From the standard normal distribution, \(P(Z > 2) = 1 - P(Z < 2) = 1 - 0.9772 = 0.0228\).For five independent selections, the probability all five women exceed 68 inches is \((0.0228)^5 \approx 6.67 \times 10^{-8}\).

Key Concepts

Standard Deviation

Standard deviation is a key concept in statistics that measures the amount of variation or dispersion in a set of values. When dealing with a normal distribution, it tells us how much individual data points tend to deviate from the mean of the dataset. In the exercise you are studying, the standard deviation of women's height is given as 2 inches. This means most women's heights will be within 2 inches of the average height of 64 inches.

• It reflects consistency or variability in a dataset.
• In a normal distribution, about 68% of data falls within one standard deviation of the mean.
• The greater the standard deviation, the more spread out the data.
  

Understanding standard deviation helps you predict how likely a data point is to be close to the average. It is foundational when calculating probabilities related to the normal distribution, as seen in parts of your exercise.

Quartiles

Quartiles divide a dataset into four equal parts. In the context of a normal distribution, quartiles help identify cutoffs for various portions of the distribution. Specifically, the first quartile (Q1) is the 25th percentile and the third quartile (Q3) is the 75th percentile. This gives us insight into the spread and skewness of the dataset.

• Q1 and Q3 reflect the values below which 25% and 75% of the data fall, respectively.
• They help in identifying the interquartile range (IQR), which is a measure of statistical dispersion and is calculated as Q3 - Q1.
• Quartiles can be used to detect outliers when data points fall significantly outside the IQR.
  

In the solution provided, Q1 and Q3 were found using z-scores, which helped translate the theoretical positions into actual height measurements, 62.652 inches and 65.348 inches respectively.

Z-Score

A z-score measures how many standard deviations an element is from the mean. Calculating z-scores allows us to standardize a normal distribution, converting different data points into a common format. This standardization enables comparisons across different datasets and simplifies probability calculations.

• Z-scores are important for determining probabilities in standard normal distributions.
• A z-score of 0 indicates a data point is exactly at the mean.
• Positive z-scores mean the data point is above the mean, while negative z-scores indicate it is below.
  

In the exercise, converting heights into z-scores allowed for the use of standard normal distribution tables, helping to find probabilities and determine quartiles effectively.

Probability Calculation

Calculating probability in a normal distribution often involves using z-scores and standard normal distribution tables. In probability theory, this helps ascertain the likelihood of an event occurring within a specified range of a dataset. For instance, the probability of a woman's height falling between 58 and 70 inches was calculated by converting these heights to their respective z-scores and finding the corresponding probability from the standard normal distribution.

• The cumulative probability helps in determining the likelihood of a random variable falling below a certain threshold.
• In the exercise, probability calculations required translating real-world measurements into z-scores.
• Tools like statistical tables or software help in obtaining these probabilities efficiently.
  

This systematic approach to probability calculation is foundational to inferential statistics, allowing predictions to be made about a population based on sample data.