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Babies in the United States have a mean birth length of \(20.5\) inches with a standard deviation of \(0.90\) inch. The shape of the distribution of birth lengths is approximately Normal. a. How long is a baby born at the 20 th percentile? b. How long is a baby born at the 50 th percentile? c. How does your answer to part b compare to the mean birth length? Why should you have expected this?

Short Answer

Expert verified
A baby born at the 20th percentile is approximately 19.74 inches long. A baby born at the 50th percentile is exactly 20.5 inches long, which also matches the mean birth length as expected in a Normal distribution.

Step by step solution

01

Calculate the 20th percentile length

Firstly, let us find the Z-Score for the 20th percentile. The Z-score corresponding to the 20th percentile is approximately -0.84. To get the actual length, we convert this Z-score into the actual length in inches by using the formula \(X = μ + Zσ\), where X is the value we want to find, μ is the mean, Z is the Z-score, and σ is the standard deviation. Hence, \(X = 20.5 + (-0.84)*0.90 = 19.744 inches.\)
02

Calculate the 50th percentile length

Now we need to calculate the 50th percentile length. The Z-score corresponding to the 50th percentile (which is the median) is 0 (as it lies in the middle of the distribution). Hence, we substitute Z=0 in the formula \(X = μ + Zσ\), and get \(X = 20.5 + 0*0.90 = 20.5 inches.\)
03

Compare the result from step 2 with the mean and analyze

Next, let's compare the length at the 50th percentile (from step 2) with the mean length. Both are exactly equal, which is expected, because for a Normal Distribution, the mean, median (50th percentile), and mode are all equal.

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Key Concepts

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

Z-Score
Understanding the Z-Score is essential for interpreting data points within a normal distribution. A Z-Score, also known as a standard score, indicates how many standard deviations an element is from the mean. It is a way to standardize scores on different scales to a common scale, often called the standard normal distribution.

This concept is used to identify where a particular score lies in relation to the average. For example, a Z-score of 0 indicates that the score is exactly at the mean, while a negative or positive Z-score indicates that the score is below or above the mean, respectively. In the context of birth lengths, a Z-score helps us identify what percentage of babies are shorter or longer than a specific length. When we calculated the 20th percentile length for babies, we used the Z-score to find out how long a baby is at the lower side of the distribution curve.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean (also known as the expected value) of the set, while a high standard deviation indicates that the values are spread out over a wider range.

In terms of the birth length example, the standard deviation tells us how much variation there is from the average (mean) birth length of 20.5 inches. The equation to find a particular value given the Z-score incorporates the standard deviation (\(X = \text{mean} + (Z \times \text{standard deviation})\)). This formula translates a Z-score back into the original score (in inches for birth length), which allows us to see how far and in what direction individual babies' lengths deviate from the average.
Normal Distribution Statistics
Normal distribution statistics are widely used to describe real-world data that tend to cluster around a mean. The normal distribution, which is also referred to as the 'bell curve', has a specific shape that is symmetrical about the mean, meaning that data near the mean are more frequent in occurrence than data far from the mean.

In the given exercise, we learn that babies' birth lengths are normally distributed, meaning if we plot these lengths, most would cluster around the average of 20.5 inches, with fewer babies being either much shorter or much longer. This pattern is crucial when we interpret percentiles. For instance, the 50th percentile, also known as the median, falls right at the mean in a normal distribution, indicating that half of the babies are longer than 20.5 inches and half are shorter. An understanding of normal distribution statistics explains why the answer to part b in the exercise is identical to the mean birth length.

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Most popular questions from this chapter

Length of Pregnancy Assume that the lengths of pregnancy for humans is approximately Normally distributed, with a mean of 267 days and a standard deviation of 10 days. Use the Empirical Rule to answer the following questions. Do not use the technology or the Normal table. Begin by labeling the horizontal axis of the graph with lengths, using the given mean and standard deviation. Three of the entries are done for you. a. Roughly what percentage of pregnancies last more than 267 days? i. almost all iii. \(68 \%\) ii. \(95 \%\) iv. \(50 \%\) b. Roughly what percentage of pregnancies last between 267 and 277 days? i. \(34 \%\) iii. \(2.5 \%\) ii. \(17 \%\) iv. \(50 \%\) c. Roughly what percentage of pregnancies last less than 237 days? i. almost all iii. \(34 \%\) ii. \(50 \%\) iv. about \(0 \%\) d. Roughly what percentage of pregnancies last between 247 and 287 days? i. almost all iii. \(68 \%\) ii. \(95 \%\) iv. \(50 \%\) e. Roughly what percentage of pregnancies last longer than 287 days? i. \(34 \%\) iii. \(2.5 \%\) ii. \(17 \%\) iv. \(50 \%\) f. Roughly what percentage of pregnancies last longer than 297 days? i. almost all iii. \(34 \%\) ii. \(50 \%\) iv. about \(0 \%\)

According to the American Veterinary Medical Association, \(30 \%\) of Americans own a cat. a. Find the probability that exactly 2 out of 8 randomly selected Americans own a cat. b. In a random sample of 8 Americans, find the probability that more than 3 own a cat.

Scores on the 2017 MCAT, an exam required for all medical school applicants, were approximately Normal with a mean score of 505 and a standard deviation of \(9.4\). a. Suppose an applicant had an MCAT score of 520 . What percentile corresponds with this score? b. Suppose to be considered at a highly selective medical school an applicant should score in the top \(10 \%\) of all test takers. What score would place an applicant in the top \(10 \%\) ?

The distribution of spring high temperatures in Los Angeles is approximately Normal, with a mean of 75 degrees and a standard deviation of \(2.5\) degrees. a. What is the probability that the high temperature is less than 70 degrees in Los Angeles on a day in spring? b. What percentage of Spring day in Los Angeles have high temperatures between 70 and 75 degrees? c. Suppose the hottest spring day in Los Angeles had a high temperature of 91 degrees. Would this be considered unusually high, given the mean and the standard deviation of the distribution? Why or why not?

The Normal model \(N(69,3)\) describes the distribution of male heights in the United States. Which of the following questions asks for a probability, and which asks for a measurement? Identify the type of problem and then answer the given question. See page 316 for guidance. a. To be a member of the Tall Club of Silicon Valley a man must be at least 74 inches tall. What percentage of men would qualify for membership in this club? b. Suppose the Tall Club of Silicon Valley wanted to admit the tallest \(2 \%\) of men. What minimum height requirement should the club set for its membership criteria?

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