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Boys Heights Heights of 10-year-old boys (5th graders) follow an approximate normal distribution with mean \(\mu=55.5\) inches and standard deviation \(\sigma=2.7\) inches \({ }^{6}\) (a) Draw a sketch of this normal distribution and label at least three points on the horizontal axis. (b) According to this normal distribution, what proportion of 10 -year-old boys are between \(4 \mathrm{ft} 4\) in and \(5 \mathrm{ft}\) tall (between 52 inches and 60 inches)? (c) A parent says his 10 -year-old son is in the 99 th percentile in height. How tall is this boy?

Short Answer

Expert verified
For question (a), the mean on the graph is marked at 55.5 inches, and the additional points that show one standard deviation from the mean are 55.5 - 2.7 = 52.8 inches and 55.5 + 2.7 = 58.2 inches. For question (b), the proportion or percentile of boys between 4ft 4in (52 inches) and 5ft (60 inches) can be found using the z-scores and the z-table. For question (c), the height of the boy at the 99th percentile can be calculated from the given percentile using the z-score and the formula \(X = μ + Zσ\).

Step by step solution

01

Sketching the Normal Distribution

Firstly, draw a symmetric bell-shaped curve to represent the normal distribution of boys' heights. The center of the curve, corresponding to the highest point, represents the mean height of 55.5 inches. Mark this point on the horizontal axis. Additionally, mark points that represent one standard deviation away from the mean on either side (55.5 - 2.7 and 55.5 + 2.7). This will give an idea of the distribution of the heights.
02

Calculating the Proportion Between Certain Heights

To find the proportion of boys between 4ft 4in and 5ft tall (or between 52 inches and 60 inches), we need to convert these values into z-scores which gives the number of standard deviations away from the mean. The formula for the z-score is \(Z = (X - μ)/σ\), where X is the value, μ is the mean and σ is the standard deviation. Calculate the z-scores for 52 and 60 inches, and then use a z-table to find the proportion of boys' heights between these two z-scores.
03

Determining the Height at the 99th Percentile

To find the height that corresponds to the 99th percentile, we first need to find the z-score that corresponds to this percentile from the z-table. This z-score gives the number of standard deviations away from the mean at 99th percentile. By using the z-score formula, we can solve for X to find the height, i.e., \(X = μ + Zσ\). Thus, the height of the boy at the 99th percentile can be found.

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Key Concepts

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

Understanding Z-Score
The z-score is a way to understand how far a specific value is from the mean of a distribution. In simple terms, it tells you how many standard deviations a value is away from the average.
To calculate a z-score, you can use the formula:
  • \( Z = \frac{(X - \mu)}{\sigma} \)
Here, \( X \) is the value you're interested in, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
A positive z-score means the value is above the mean, and a negative z-score means it's below. Around 68% of data falls within one standard deviation, so most z-scores are between -1 and 1.
Z-scores help us use the z-table to find probabilities, showing what percent of data lies below or between certain values. This is how we understand the likelihood of an event within a given distribution.
Demystifying Percentiles
Percentiles rank the data in a distribution, giving us an idea of how a specific value compares to the rest. For instance, being in the 99th percentile means you score higher than 99% of the other values.
In terms of height, if a boy is in the 99th percentile, he is taller than 99% of his peers. To find which height corresponds to a specific percentile, look up the z-score for that percentile and use the formula:
  • \( X = \mu + Z\sigma \)
This will give you the actual value, like height, associated with the percentile. Percentiles are useful because they provide insights into data's relative standing.
Explaining Standard Deviation
Standard deviation is a measure of how spread out the numbers in a data set are. It's a key component in the normal distribution and helps to quantify the amount of variation.
When you have a small standard deviation, the data points are clustered closely around the mean. A larger standard deviation means they are spread out over a wider range of values.
In a normal distribution, approximately 68% of data points lie within one standard deviation of the mean, 95% within two, and 99.7% within three.
In our example, with a mean height of 55.5 inches and a standard deviation of 2.7 inches, most 10-year-old boys’ heights would fall between 52.8 inches and 58.2 inches.
Understanding standard deviation helps us comprehend the diversity in data, enabling us to predict probability and variability in real-world scenarios.

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Most popular questions from this chapter

Curving Grades on an Exam A statistics instructor designed an exam so that the grades would be roughly normally distributed with mean \(\mu=75\) and standard deviation \(\sigma=10 .\) Unfortunately, a fire alarm with ten minutes to go in the exam made it difficult for some students to finish. When the instructor graded the exams, he found they were roughly normally distributed, but the mean grade was 62 and the standard deviation was 18\. To be fair, he decides to "curve" the scores to match the desired \(N(75,10)\) distribution. To do this, he standardizes the actual scores to \(z\) -scores using the \(N(62,18)\) distribution and then "unstandardizes" those \(z\) -scores to shift to \(N(75,10)\). What is the new grade assigned for a student whose original score was \(47 ?\) How about a student who originally scores a \(90 ?\)

(a) The area below \(z=1.04\) (b) The area above \(z=-1.5\) (c) The area between \(z=1\) and \(z=2\)

Heights of Men in the US Heights of adult males in the US are approximately normally distributed with mean 70 inches \((5 \mathrm{ft} 10 \mathrm{in})\) and standard deviation 3 inches. (a) What proportion of US men are between \(5 \mathrm{ft}\) 8 in and \(6 \mathrm{ft}\) tall \((68\) and 72 inches, respectively)? (b) If a man is at the 10 th percentile in height, how tall is he?

Hearing Loss in Teenagers A recent study" found that, of the 1771 participants aged 12 to 19 in the National Health and Nutrition Examination Survey, \(19.5 \%\) had some hearing loss (defined as a loss of 15 decibels in at least one ear). This is a dramatic increase from a decade ago. The sample size is large enough to use the normal distribution, and a bootstrap distribution shows that the standard error for the proportion is \(S E=0.009 .\) Find and interpret a \(90 \%\) confidence interval for the proportion of teenagers with some hearing loss.

Exam Grades Exam grades across all sections of introductory statistics at a large university are approximately normally distributed with a mean of 72 and a standard deviation of \(11 .\) Use the normal distribution to answer the following questions. (a) What percent of students scored above a \(90 ?\) (b) What percent of students scored below a \(60 ?\) (c) If the lowest \(5 \%\) of students will be required to attend peer tutoring sessions, what grade is the cutoff for being required to attend these sessions? (d) If the highest \(10 \%\) of students will be given a grade of \(\mathrm{A},\) what is the cutoff to get an \(\mathrm{A}\) ?

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