Question

How do you survive when there's no time to eat-fast food, no food, a protein bar, candy? A Snapshot in USA Today indicates that \(36 \%\) of women aged \(25-55\) say that, when they are too busy to eat, they get fast food from a drive-thru. \({ }^{13}\) A random sample of 100 women aged \(25-55\) is selected. a. What is the average number of women who say they eat fast food when they're too busy to eat? b. What is the standard deviation for the number of women who say they eat fast food when they're too busy to eat? c. If 49 of the women in the sample said they eat fast food when they're too busy to eat, would this be an unusual occurrence? Explain.

Short answer

Answer: Yes, it is unusual.

Step-by-step solution

a. Finding the average number of women who say they eat fast food when they're too busy to eat

To find the average number of women, we can multiply the percentage by the sample size, which is 100 in this case. Average = (Percentage of women who eat fast food) × (Sample size) Average = (0.36) × (100) Average = 36 So, the average number of women who say they eat fast food when they're too busy to eat is 36.

b. Finding the standard deviation for the number of women who say they eat fast food when they're too busy to eat

To find the standard deviation, we can use the formula for the standard deviation of a binomial distribution: Standard deviation = \( \sqrt{n \times p \times (1-p)} \) Where n is the sample size (100 in this case), p is the percentage of women who eat fast food (0.36), and (1-p) is the percentage of women who don't eat fast food (0.64). Standard deviation = \( \sqrt{100 \times 0.36 \times 0.64} \) Standard deviation ≈ 4.74 So, the standard deviation for the number of women who say they eat fast food when they're too busy to eat is approximately 4.74.

c. Determining if 49 women in the sample eating fast food when they're too busy to eat is an unusual occurrence

To determine if 49 women in the sample is unusual, we can calculate the z-score with the following formula: Z-score = \( \frac{x - \mu}{\sigma} \) Where x is the number of women in the sample eating fast food when too busy (49), μ is the average (36), and σ is the standard deviation (4.74). Z-score = \( \frac{49 - 36}{4.74} \) Z-score ≈ 2.74 A z-score greater than 2 or less than -2 is generally considered to be unusual. In this case, the z-score is approximately 2.74, which is greater than 2. Hence, having 49 women in the sample say they eat fast food when they're too busy to eat would indeed be an unusual occurrence.

Key concepts

Standard Deviation

In the realm of probability and statistics, standard deviation is key to understanding data variability. When we talk about standard deviation, we refer to a measure that shows how much the values in a data set differ from the average (mean) value. In simple terms, it helps us understand how spread out the numbers in a data set are. The lower the standard deviation, the more closely the data points cluster around the mean; conversely, a higher standard deviation indicates greater dispersion.

For example, in our binomial distribution exercise concerning women who eat fast food because they’re too busy, the standard deviation helps determine how much the number of women who do is likely to vary. This variance is captured using the formula for the standard deviation of a binomial distribution:

• Standard deviation = \( \sqrt{n \times p \times (1-p)} \)
• Where \(n\) is the sample size, \(p\) is the probability of success (e.g., the fraction of women opting for fast food), and \((1-p)\) is the probability of failure.

The computed standard deviation of approximately 4.74 indicates how much variance we can expect in different random samples of 100 women.

Z-Score

Z-score is a statistical concept that relates a data point to the mean and standard deviation of a data set. It helps to determine how far or close an individual data point is from the mean. Expressed in terms of standard deviations, Z-score is essential in identifying how unusual or typical a data occurrence is.

In layman's terms, it's a way to place a score into a normal distribution and see where it stands compared to the average.

In our context, to decide if finding 49 women in a sample who eat fast food because they are too busy is unusual, we calculate the Z-score using the formula:

• Z-score = \( \frac{x - \mu}{\sigma} \)
• Where \(x\) is the number of women in this particular instance (49), \(\mu\) is the average (36), and \(\sigma\) is the standard deviation (4.74).

With a Z-score of approximately 2.74, we find that this value exceeds the common threshold of 2 in absolute terms, designating this result as an unusual occurrence in a statistical sense.

Random Sample

In statistics, a random sample is a subset of individuals from a larger population, chosen in such a way that every individual had an equal chance of being selected. The concept essentially ensures fairness and randomness in the selection process, eliminating bias.

In our given exercise, a random sample of 100 women is chosen from the population of women aged 25-55. This selection ensures that the insights, like the tendency to opt for fast food when busy, are representative not of biases but of the genuine behavior pattern within the whole population.

Understanding random sampling is crucial because it impacts the reliability of statistical conclusions. By grounding analyses in random samples, statisticians can infer the characteristics of entire populations without needing data from each member. This ensures data-backed conclusions remain both efficient and robust in various fields like market research, policy-making, and more.