Question

An advertiser claims that the average percentage of brown M\&M'S candies in a package of milk chocolate M\&M'S is \(13 \% .\) Suppose you randomly select a package of milk chocolate M\&M'S that contains 55 candies and determine the proportion of brown candies in the package. a. What is the approximate distribution of the sample proportion of brown candies in a package that contains 55 candies? b. What is the probability that the sample percentage of brown candies is less than \(20 \% ?\) c. What is the probability that the sample percentage exceeds \(35 \% ?\) d. Within what range would you expect the sample proportion to lie about \(95 \%\) of the time?

Short answer

Additionally, what is the range within which the sample proportion lies 95% of the time? The probability that the sample proportion of brown M&M'S candies is less than 20% is approximately 94.32%. The probability that the sample proportion exceeds 35% is 0%. The sample proportion lies within the range of approximately 4.45% to 21.55% about 95% of the time.

Step-by-step solution

Calculate the population proportion

The given average percentage of brown M&M'S candies in a package is 13%. Therefore, we have p = 0.13 (in proportion).

Finding the standard deviation of the sample proportion

We use the following formula to find the standard deviation (σ) of the sample proportion: σ = sqrt((p * (1-p)) / n) where n is the number of candies in the selected sample (55). σ = sqrt((0.13 * (1-0.13)) / 55) σ ≈ 0.0442

Calculate the z-score for given percentages

To find the z-score, we will use the formula: z = (x - μ) / σ where x is the sample proportion, μ is the population proportion, and σ is the standard deviation of the sample proportion. We will calculate z at 20% and 35% for parts b and c.

Part b: Probability that the sample percentage is less than 20%

We will first calculate the z-score at 20% (0.2 in proportion). z = (0.2 - 0.13) / 0.0442 z ≈ 1.5861 Now, by looking up in the z-table, the probability is found for z = 1.5861 P(z < 1.5861) ≈ 0.9432

Part c: Probability that the sample percentage exceeds 35%

Calculate the z-score at 35% (0.35 in proportion). z = (0.35 - 0.13) / 0.0442 z ≈ 4.9774 By looking up the z-table, we find: P(z < 4.9774) ≈ 1 Since we want the probability of the sample to exceed 35%, we subtract the found probability from 1. P(z > 4.9774) = 1 - P(z < 4.9774) = 1 - 1 = 0

Part d: Range within which the sample proportion lies 95% of the time

Since we are given a confidence level of 95%, we will look for z-scores corresponding to the 2.5% and 97.5% percentiles in the z-table, which are -1.96 and 1.96, respectively. To find the range, we'll use the formula: Sample proportion = μ ± (z * σ) Lower limit: 0.13 - (1.96 * 0.0442) ≈ 0.0445 Upper limit: 0.13 + (1.96 * 0.0442) ≈ 0.2155 Thus, within the range of approximately 4.45% to 21.55%, we can expect the sample proportion to lie about 95% of the time.

Key concepts

Population Proportion

In statistics, the population proportion refers to the fraction of members in a group displaying a specific characteristic. In our scenario, this characteristic is being a brown M&M. The population proportion is usually denoted as \( p \). For example, if an advertiser claims that 13% of M&M's in a package are brown, we can write \( p = 0.13 \). This is crucial for analyzing the sample data, as it serves as our expected value to which sample data can be compared.

Standard Deviation

Standard deviation measures the spread or variability of a data set. In context, it assesses how much the observed proportions in different samples are likely to vary from the population proportion. To calculate the standard deviation of our sample proportion, we use the formula:\[ \sigma = \sqrt{\frac{p(1-p)}{n}} \] where \( p \) is the population proportion, and \( n \) is the sample size. For our M&M's example, with \( p = 0.13 \) and \( n = 55 \), the calculation becomes:\[ \sigma = \sqrt{\frac{0.13 \times (1-0.13)}{55}} \approx 0.0442 \]This tells us how much the sample proportion varies from the population proportion, giving us a sense of reliability in our estimation.

Z-score

The Z-score is a statistical measure that describes a value's relationship to the mean of a group of values. It is measured in terms of standard deviations from the mean. For probabilities, the Z-score helps in determining how far away a sample proportion is from the population proportion. The formula for the Z-score is:\[ z = \frac{x - \mu}{\sigma} \]where \( x \) is the sample proportion, \( \mu \) is the population proportion (i.e., mean), and \( \sigma \) is the standard deviation of the sample proportion. For the probabilities in our M&M problem, you calculate the Z-score for 20% and 35%, providing a standard way to compare these probabilities against a normal distribution.

Probability Distribution

A probability distribution describes how the values of a random variable are distributed. In this context, it represents the likelihood of different outcomes for the sample proportion of brown M&M's in a sample taken from a population. Typically, sample proportions followed a normal distribution if the sample size is sufficiently large, thanks to the Central Limit Theorem. This assumption sometimes allows us to use Z-scores to find probabilities for specific outcomes, such as the probability of encountering less than 20% or more than 35% brown M&M's in a package.

Confidence Interval

A confidence interval provides a range of values that is likely to contain a population parameter with a certain level of confidence. In our case, it's used to estimate the range within which we expect the true proportion of brown M&M's to lie 95% of the time.For a 95% confidence level, you often use the critical Z-scores of -1.96 and 1.96, which correspond to the 2.5% tails of the normal distribution. The formula to find the confidence interval is:\[ \text{Sample proportion} = \mu \pm (z \times \sigma) \]Using the M&M example, with \( \mu = 0.13 \) and \( \sigma = 0.0442 \), the interval is approximately 4.45% to 21.55%. This tells us that you can be pretty confident the true proportion of brown M&M’s is within this range.