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The paper referenced in Example 6.21 suggested that a normal distribution with mean 3500 grams and standard deviation 550 grams is a reasonable model for birth weights of babies born in Canada. a. One common medical definition of a large baby is any baby that weighs more than 4000 grams at birth. What is the probability that a randomly selected Canadian baby is a large baby? b. What is the probability that a randomly selected Canadian baby weighs either less than 2000 grams or more than 4000 grams at birth? c. What birth weights describe the \(10 \%\) of Canadian babies with the greatest birth weights? 6.52 The time that it takes a randomly selected job applicant to perform a certain task has a distribution that can be approximated by a normal distribution with a mean of 120 seconds and a standard deviation of 20 seconds. The fastest \(10 \%\) are to be given advanced training. What task times qualify individuals for such training?

Short Answer

Expert verified
The short version of the answer is as follows: a. The probability that a randomly selected Canadian baby is a large baby (weighs more than 4000 grams) is approximately 18.14%. b. The probability that a randomly selected Canadian baby weighs either less than 2000 grams or more than 4000 grams at birth is approximately 18.46%. c. The birth weights that describe the top 10% of Canadian babies are around 4204 grams.

Step by step solution

01

Calculate Z-scores for given weights

First, for each of the given weights, we'll calculate the Z-scores using the formula: \(Z = \frac{X - \mu}{\sigma}\), where \(X\) is the given weight, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. For 4000 grams: \(Z_1 = \frac{4000 - 3500}{550} = 0.91\) For 2000 grams: \(Z_2 = \frac{2000 - 3500}{550} = -2.73\)
02

Calculate the probabilities associated with the Z-scores

Using the Z-table, we'll find the probabilities associated with the Z-scores: For \(Z_1 = 0.91\), the probability is \(P(Z \leq 0.91) = 0.8186\) For \(Z_2 = -2.73\), the probability is \(P(Z \leq -2.73) = 0.0032\)
03

Answering part a

To find the probability of a baby weighing more than 4000 grams at birth, we need to find the area in the tail, which is \(1 - P(Z \leq 0.91) = 1 - 0.8186 = 0.1814\). Thus, the probability that a randomly selected Canadian baby is a large baby is 18.14%.
04

Answering part b

To find the probability of a baby weighing less than 2000 grams or more than 4000 grams, we need to find the area in both tails. We already have the probabilities for these tails, so we just need to add them: \(P(Z \leq -2.73) + (1 - P(Z \leq 0.91)) = 0.0032 + 0.1814 = 0.1846\). Thus, the probability that a randomly selected Canadian baby weighs either less than 2000 grams or more than 4000 grams at birth is 18.46%.
05

Answering part c

To find the birth weights for the top 10% of Canadian babies, we need to find the Z-score corresponding to the top 10%, or the 90th percentile. Using the Z-table, we find that the Z-score for the 90th percentile is approximately 1.28. Now, we'll use the formula for Z-score to find the corresponding weight: \(X = \mu + Z \cdot \sigma\): \(X = 3500 + 1.28 \cdot 550 = 3500 + 704 = 4204\) The birth weights that describe the top 10% of Canadian babies are around 4204 grams.

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Key Concepts

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

Z-score
The concept of the Z-score is fundamental when dealing with normal distributions. A Z-score measures how many standard deviations an element is from the mean. It provides a simple way to compare values coming from different normal distributions or to convert a score to a standard form. The formula for calculating a Z-score is:\[ Z = \frac{X - \mu}{\sigma} \]where:
  • \( X \) is the value we are examining,
  • \( \mu \) is the mean of the distribution, and
  • \( \sigma \) is the standard deviation.
  • This transformation allows us to assess how typical or atypical a particular observation is within the context of the entire dataset. For instance, if 4000 grams is the weight of a Canadian newborn, you can find how it compares to the average weight by calculating its Z-score. With the given mean (3500 grams) and standard deviation (550 grams), a weight of 4000 grams has a Z-score of 0.91, indicating it is 0.91 standard deviations above the mean. This helps us determine the likelihood of a randomly selected baby weighing this much.
Probability Calculation
Probability calculations using the normal distribution enable us to figure out how likely certain outcomes are. These calculations often involve the use of Z-tables or statistical software to determine the probability associated with a certain Z-score. Let's have a closer look at the relevant probabilities for Canadian birth weights:1. **Probability of a Large Baby**: Once we compute the Z-score for a weight over 4000 grams and find it to be 0.91, we can use a Z-table to find that \( P(Z \leq 0.91) \approx 0.8186 \). This tells us the probability that a baby weighs *up to* 4000 grams. To find the probability of a baby being larger, we compute \( 1 - 0.8186 = 0.1814 \). Hence, 18.14% of Canadian babies are considered large.2. **Probability of a Baby Outside the Normal Range (less than 2000 or more than 4000 grams)**: This kind of calculation involves probabilities from both tails of the normal distribution. Once we know \( P(Z \leq -2.73) \approx 0.0032 \), the probability of a baby weighing less than 2000 grams, we add it to the probability of weighing more than 4000 grams to get a total probability of 18.46%.
Percentiles
Percentiles are another useful concept when dealing with normal distributions. They help us understand and describe relative standing of a value in a dataset. Specifically, a percentile indicates the value below which a given percentage of observations fall.In the case of Canadian birth weights, we wanted to find the weight corresponding to the 90th percentile, indicating the top 10% of baby weights. To determine this, we looked up the Z-score that corresponds to the 90th percentile. A Z-score of about 1.28 is typical for the 90th percentile in a standard normal distribution.Using this Z-score, we calculated:\[ X = 3500 + 1.28 \times 550 \]\[ X \approx 4204 \text{ grams} \]This means that approximately 10% of Canadian babies weigh more than 4204 grams. The knowledge of where a particular weight falls in the distribution—through percentiles—helps in making informed decisions, such as providing targeted medical care.
Canadian Birth Weights
When talking about Canadian birth weights, the underlying assumption often is that they follow a normal distribution. This allows medical researchers and statisticians to make numeric probability-based insights such as the likelihood of a baby being classified as 'large' or to identify at what weights certain percentiles of the population lie. The mean weight of 3500 grams and a standard deviation of 550 grams serve as the central parameters of this distribution, providing a baseline to make probability calculations and analyze specific outcomes.
  • **Practical Applications**: Understanding these weights helps in allocating healthcare resources efficiently and in establishing guidelines such as what constitutes a low, average, or high birth weight.
  • **Use in Public Health**: It is crucial in public health planning and policy formulation ensuring that babies and mothers receive the appropriate care and support.
This perspective on Canadian birth weights offers a blend of practical insights and theoretical understanding, making it invaluable to both healthcare professionals and statisticians.

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