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Heights of 10 year olds, Part I. Heights of 10 year olds, regardless of gender, closely follow a normal distribution with mean 55 inches and standard deviation 6 inches. (a) What is the probability that a randomly chosen 10 year old is shorter than 48 inches? (b) What is the probability that a randomly chosen 10 year old is between 60 and 65 inches? (c) If the tallest \(10 \%\) of the class is considered "very tall", what is the height cutoff for "very tall"?

Short Answer

Expert verified
(a) 12.10% are shorter than 48 inches. (b) 15.58% are between 60 and 65 inches. (c) The cutoff for "very tall" is approximately 62.68 inches.

Step by step solution

01

Identify Given Values

We have a normal distribution for the heights with a mean \( \mu = 55 \) inches and a standard deviation \( \sigma = 6 \) inches.
02

Calculate Z-Score for Part (a)

For part (a), find the Z-score using the formula: \[ Z = \frac{X - \mu}{\sigma} \] where \( X = 48 \) inches.\[ Z = \frac{48 - 55}{6} = \frac{-7}{6} \approx -1.17 \]
03

Look Up Z-Score in Standard Normal Distribution Table for Part (a)

Find the probability corresponding to \( Z = -1.17 \) using a standard normal distribution table or a calculator. This gives \( P(Z < -1.17) \approx 0.1210 \).
04

Calculate Z-Scores for Part (b)

For part (b), calculate the Z-scores for both 60 and 65 inches.For 60 inches: \[ Z = \frac{60 - 55}{6} = \frac{5}{6} \approx 0.83 \] For 65 inches:\[ Z = \frac{65 - 55}{6} = \frac{10}{6} \approx 1.67 \]
05

Look Up Z-Scores in Standard Normal Distribution Table for Part (b)

Find the probabilities for each Z-score:- \( P(Z < 0.83) \approx 0.7967 \)- \( P(Z < 1.67) \approx 0.9525 \)Find the probability between 60 and 65 inches: \( P(0.83 < Z < 1.67) = P(Z < 1.67) - P(Z < 0.83) \approx 0.9525 - 0.7967 = 0.1558 \).
06

Find Z-Score for Tallest 10% for Part (c)

We want the height cutoff above which lies the tallest 10% of the class. Look for the Z-score corresponding to the 90th percentile, which is about \( Z \approx 1.28 \) from the Z-table.
07

Calculate Height Cutoff Using Z-Score for Part (c)

Using the Z-score \( Z = 1.28 \), solve for \( X \) using the formula:\[ X = Z \sigma + \mu \]\[ X = 1.28 \times 6 + 55 = 7.68 + 55 = 62.68 \]Therefore, the height cutoff for "very tall" is approximately 62.68 inches.

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Key Concepts

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

Standard Deviation
In a normal distribution, the standard deviation is a measure of how spread out the values are around the mean. It's denoted by the symbol \( \sigma \) (sigma) and helps us understand the variability within a set of data. For example, in our exercise, the standard deviation is 6 inches. This means that most of the heights are within 6 inches of the average height, which is 55 inches.

A smaller standard deviation would mean the heights are closer to the mean, indicating less variability. A larger one signifies more spread out data, indicating greater variability. This concept is key when we are interpreting statistical data because it gives context to the mean.

Understanding standard deviation is essential for making predictions about the probability of various outcomes and is widely used across different fields, including psychology, economics, and even weather forecasting.
Z-Score
A Z-score helps us understand how far and in what direction a data point is from the mean, measured in standard deviations. The formula for calculating a Z-score is \( Z = \frac{X - \mu}{\sigma} \). Here, \( X \) is the value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

For instance, if a 10-year-old child is 48 inches tall, the Z-score calculation would be \( Z = \frac{48 - 55}{6} \), which equals approximately -1.17. This indicates the child's height is 1.17 standard deviations below the mean. Z-scores are helpful because they allow us to determine the position of individual data points within a distribution and to compare scores from different normal distributions.

Interpreting Z-scores helps in understanding where an individual stands compared to the average population, making it a valuable tool in statistics and research.
Probability Calculation
Calculating probabilities in a normal distribution involves using Z-scores to find the likelihood of a particular event. The normal distribution table, also known as the Z-table, provides the probability that a statistic is less than a given Z-score.

In the exercise, to find the probability of being shorter than 48 inches, the Z-score was calculated to be approximately -1.17. Using the Z-table, we found the probability \( P(Z < -1.17) \) to be about 0.1210, indicating that there's a 12.10% chance of randomly selecting a child shorter than 48 inches.

For heights between 60 and 65 inches, we calculate probabilities for Z-scores of 0.83 and 1.67 respectively. These help us find the probability that a randomly selected height falls within this range. Such calculations are crucial for making inferences about a population, assessing risk, and decision-making in various fields.
Percentile
Percentiles in a normal distribution indicate the percentage of data below a specific value. Percentiles are useful for comparing individual scores with the entire data set to understand how a particular score ranks.

In the given exercise, we needed to determine the height cutoff for the tallest 10% of children. This required finding the 90th percentile, which corresponds to a Z-score of about 1.28. Using this Z-score, we calculated the height cutoff to be approximately 62.68 inches. This means 90% of children are shorter than 62.68 inches, and the tallest 10% are above this height.

Percentiles are extensively used in testing and assessments to interpret individual scores and set benchmarks for performance comparisons, making them an essential component of statistical analysis.
Statistical Distribution
A statistical distribution describes all the possible values and frequencies of those values for a given data set. One of the most well-known statistical distributions is the normal distribution, which is symmetrical around its mean, representing data that tend to cluster around a central value.

In our exercise, the heights of 10-year-olds follow a normal distribution, characterized by a mean of 55 inches and a standard deviation of 6 inches. Normal distributions are significant in statistics because many variables observed in nature and society, such as heights and test scores, loosely follow this pattern.

Understanding statistical distributions allows us to model real-world phenomena, make predictions, and develop strategies for various applications, including quality control, finance, and scientific research. Recognizing the type of distribution that best represents your data is the first step in performing more advanced statistical analysis.

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Most popular questions from this chapter

Exercise 4.36 states that the distribution of speeds of cars traveling on the Interstate 5 Freeway (I-5) in California is nearly normal with a mean of 72.6 miles/hour and a standard deviation of 4.78 miles/hour. The speed limit on this stretch of the I-5 is 70 miles/hour. (a) A highway patrol officer is hidden on the side of the freeway. What is the probability that 5 cars pass and none are speeding? Assume that the speeds of the cars are independent of each other. (b) On average, how many cars would the highway patrol officer expect to watch until the first car that is speeding? What is the standard deviation of the number of cars he would expect to watch?

How many cars show up? For Monday through Thursday when there isn't a holiday, the average number of vehicles that visit a particular retailer between \(2 \mathrm{pm}\) and \(3 \mathrm{pm}\) each afternoon is \(6.5,\) and the number of cars that show up on any given day follows a Poisson distribution. (a) What is the probability that exactly 5 cars will show up next Monday? (b) What is the probability that \(0,1,\) or 2 cars will show up next Monday between \(2 \mathrm{pm}\) and \(3 \mathrm{pm} ?\) (c) There is an average of 11.7 people who visit during those same hours from vehicles. Is it likely that the number of people visiting by car during this hour is also Poisson? Explain.

Customers at a coffee shop. A coffee shop serves an average of 75 customers per hour during the morning rush. (a) Which distribution have we studied that is most appropriate for calculating the probability of a given number of customers arriving within one hour during this time of day? (b) What are the mean and the standard deviation of the number of customers this coffee shop serves in one hour during this time of day? (c) Would it be considered unusually low if only 60 customers showed up to this coffee shop in one hour during this time of day? (d) Calculate the probability that this coffee shop serves 70 customers in one hour during this time of day.

Playing darts. Calculate the following probabilities and indicate which probability distribution model is appropriate in each case. A very good darts player can hit the bull's eye (red circle in the center of the dart board) \(65 \%\) of the time. What is the probability that he (a) hits the bullseye for the \(10^{\text {th }}\) time on the \(15^{\text {th }}\) try? (b) hits the bullseye 10 times in 15 tries? (c) hits the first bullseye on the third try?

With and without replacement. In the following situations assume that half of the specified population is male and the other half is female. (a) Suppose you're sampling from a room with 10 people. What is the probability of sampling two females in a row when sampling with replacement? What is the probability when sampling without replacement? (b) Now suppose you're sampling from a stadium with 10,000 people. What is the probability of sampling two females in a row when sampling with replacement? What is the probability when sampling without replacement? (c) We often treat individuals who are sampled from a large population as independent. Using your findings from parts (a) and (b), explain whether or not this assumption is reasonable.

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