/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 According to National Vital Stat... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

According to National Vital Statistics, the average length of a newborn baby is \(19.5\) inches with a standard deviation of \(0.9\) inches. The distribution of lengths is approximately Normal. Use technology or a table to answer these questions. For each include an appropriately labeled and shaded Normal curve. a. What is the probability that a newborn baby will have a length of 18 inches or less? b. What percentage of newborn babies will be longer than 20 inches? c. Baby clothes are sold in a "newborn" size that fits infants who are between 18 and 21 inches long. What percentage of newborn babies will not fit into the "newborn" size either because they are too long or too short?

Short Answer

Expert verified
a) The probability that a newborn baby will have a length of 18 inches or less is approximately 4.75%. b) About 28.75% of newborn babies will be longer than 20 inches. c) Approximately 9.5% of newborn babies will not fit into the 'newborn' size either because they are too long (longer than 21 inches) or too short (shorter than 18 inches).

Step by step solution

01

Calculating the Z-score for a length of 18 inches

The Z-score is calculated using the formula \(Z = (X - \mu) / \sigma\), where X is the value from the data, \(\mu\) is the mean and \(\sigma\) is the standard deviation. Substituting the values, we get \(Z = (18 - 19.5) / 0.9 = -1.67\). This Z-score tells us that 18 inches is 1.67 standard deviations below the mean.
02

Finding the probability that a newborn baby will have a length of 18 inches or less

This is equivalent to finding the probability for a Z-score of -1.67 or lower. Looking up this Z-score in the standard Normal Distribution table (or using a calculator or software), we find that the probability is around 0.0475 or 4.75%.
03

Calculating the Z-score for a length of 20 inches

Again, using the Z-score formula \(Z = (X - \mu) / \sigma\), we find that the Z-score for 20 inches is \(Z = (20 - 19.5) / 0.9 = 0.56\). This Z-score tells us that 20 inches is 0.56 standard deviations above the mean.
04

Finding the percentage of newborn babies longer than 20 inches

This is equivalent to finding the probability for a Z-score of 0.56 or higher. Looking up this Z-score in the standard Normal Distribution table, we find that the probabilty for a Z-score lower than 0.56 is around 0.7125. So the percentage of babies longer than 20 inches is 1 - 0.7125 = 0.2875 or 28.75%.
05

Calculating the Z-score for a length of 21 inches and finding the percentage of newborn babies not fitting in 'newborn' size

First, we find the Z-score for 21 inches. Using the Z-score formula again, we find that the Z-score for 21 inches is \(Z = (21 - 19.5) / 0.9 = 1.67\). Next, we look up this Z-score in the table and find that the probability for a Z-score lower than 1.67 is 0.9525. So the percentage of babies shorter than 18 inches or longer than 21 inches is (1 - 0.9525) + 0.0475 = 0.095 or 9.5%.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Z-score Calculation
Understanding the Z-score is crucial for deciphering the normal distribution. It's a measure that indicates how many standard deviations a particular value is from the mean. To calculate the Z-score, you use the formula:
\[ Z = \frac{(X - \mu)}{\sigma} \]
where \(X\) is the value you're interested in, \(\mu\) represents the mean of the dataset, and \(\sigma\) is the standard deviation.

In our exercise, when calculating the Z-score for an 18-inch baby, we find the baby's length is 1.67 standard deviations below the average since the Z-score is -1.67. This helps us understand where the baby’s length falls in relation to the rest of the population. For instance, a Z-score of -1.67 corresponds to a low percentile, implying such a length is rare. Collectively, grasping how to compute and interpret Z-scores is indispensable for analyzing data using the standard normal distribution.
Standard Normal Distribution
The standard normal distribution, also known as the Z-distribution, is a normal distribution with a mean of zero and a standard deviation of one. It's used as a reference point to understand how uncommon or common a particular value is within a normal distribution.

Once you have the Z-score, you can refer to standard normal distribution tables or use statistical software to find probabilities. For instance, in the exercise, after calculating a Z-score for 18 inches (-1.67), we consulted a standard normal distribution table to determine the probability of a baby being that length or shorter, which turned out to be around 4.75%. This process is a vital part of statistical analysis and probability, especially in fields like psychology, economics, and biology where normal distributions frequently occur.
Normal Curve
The normal curve, also known as the bell curve, is a graphical representation of the normal distribution. It's symmetrical, with most of the observations clustering around the mean and probabilities trailing off towards the extremes.

When working with data like the lengths of newborn babies, plotting these lengths on a normal curve helps to visualize the distribution and quickly identify the proportions lying within certain intervals. The exercise illustrates this as it asks for appropriately labeled and shaded normal curves representing different scenarios.
  • For lengths 18 inches or less, the left tail of the normal curve is shaded representing 4.75%.
  • For those longer than 20 inches, the right tail is shaded to signify 28.75%.
  • Finally, for the range not fitting into 'newborn' size, two outer segments are shaded to depict 9.5% of the total.
Understanding the shape and properties of the normal curve is essential in statistics as it allows for better interpretation of data and probabilities.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The distribution of red blood cell counts is different for men and women. For both, the distribution is approximately Normal. For men, the middle \(95 \%\) range from \(4.5\) to \(5.7\) million cells per microliter and for women, the middle \(95 \%\) have red blood cells counts between \(3.9\) and \(5.0\) million cells per microliter. a. What is the mean and the standard deviation of red blood cell counts for men? Explain your reasoning. b. What is the mean and the standard deviation of red blood cell counts for women? Explain your reasoning. c. Which gender shows more variation in red blood cell counts? Support your answer with appropriate statistics.

According to dogtime .com, the mean weight of an adult St. Bernard dog is 150 pounds. Assume the distribution of weights is Normal with a standard deviation of 10 pounds. a. Find the standard score associated with a weight of 170 pounds. b. Using the Empirical Rule and your answer to part a, what is the probability that a randomly selected St. Bernard weighs more than 170 pounds? c. Use technology to confirm your answer to part \(\mathrm{b}\) is correct. d. Almost all adult St. Bernard's will have weights between what two values?

Quantitative SAT scores are approximately Normally distributed with a mean of 500 and a standard deviation of \(100 .\) On the horizontal axis of the graph, indicate the SAT scores that correspond with the provided \(z\) -scores. (See the labeling in Exercise 6.14.) Answer the questions using only your knowledge of the Empirical Rule and symmetry. a. Roughly what percentage of students earn quantitative SAT scores greater than \(500 ?\) i. almost all iii. \(50 \%\) \(\mathrm{v}\). about \(0 \%\) ii. \(75 \%\) iv. \(25 \%\) b. Roughly what percentage of students earn quantitative SAT scores between 400 and \(600 ?\) i. almost all iii. \(68 \%\) \(\mathrm{v}\). about \(0 \%\) ii. \(95 \%\) iv. \(34 \%\) c. Roughly what percentage of students earn quantitative SAT scores greater than \(800 ?\) i. almost all iii. \(68 \%\) \(\mathrm{v}\). about \(0 \%\) ii. \(95 \%\) iv. \(34 \%\) d. Roughly what percentage of students earn quantitative SAT scores les: than \(200 ?\) i. almost all iii. \(68 \%\) \(\mathrm{v}\). about \(0 \%\) ii. \(95 \%\) iv. \(34 \%\) e. Roughly what percentage of students earn quantitative SAT scores between 300 and \(700 ?\) i. almost all iii. \(68 \%\) v. \(2.5 \%\) ii. \(95 \%\) iv. \(34 \%\) f. Roughly what percentage of students earn quantitative SAT scores between 700 and 800 ? i. almost all iii. \(68 \%\) v. \(2.5 \%\) ii. \(95 \%\) iv. \(34 \%\)

The distribution of grade point averages GPAs for medical school applicants in 2017 were approximately Normal, with a mean of \(3.56\) and a standard deviation of \(0.34\). Suppose a medical school will only consider candidates with GPAs in the top \(15 \%\) of the applicant pool. An applicant has a GPA of \(3.71\). Does this GPA fall in the top \(15 \%\) of the applicant pool?

Assume a standard Normal distribution. Draw a separate, well-labeled Normal curve for each part. a. Find an approximate \(z\) -score that gives a left area of \(0.7000\). b. Find an approximate \(z\) -score that gives a left area of \(0.9500\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.