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Chapter 2: Problem 53

Length of pregnancies The length of human pregnancies from conception to birth varies according to a distribution that is approximately Normal with mean 266 days and standard deviation 16 days. For each part, follow the four-step process. (a) At what percentile is a pregnancy that lasts 240 days (that鈥檚 about 8 months)? (b) What percent of pregnancies last between 240 and 270 days (roughly between 8 months and 9 months)? (c) How long do the longest 20% of pregnancies last?

Short Answer

Expert verified
(a) 5.2nd percentile. (b) 54.67% of pregnancies. (c) 280 days or more.

Step by step solution

01

Understanding the Problem

We are given a normal distribution for pregnancy lengths with a mean (\(\mu\)) of 266 days and a standard deviation (\(\sigma\)) of 16 days. We need to find the percentile rank of a 240-day pregnancy, the percentage of pregnancies lasting between 240 and 270 days, and the cutoff duration for the longest 20% of pregnancies.
02

Convert Days to Z-scores for Part (a)

To find the percentile rank for a pregnancy lasting 240 days, we first convert 240 days into a z-score using the formula: \[ z = \frac{x - \mu}{\sigma} \]. For \(x = 240\), \(z = \frac{240 - 266}{16} = \frac{-26}{16} = -1.625\).
03

Find the Percentile for Part (a)

Using the z-score of -1.625, we consult a standard normal distribution table or calculator to find the percentile rank. A z-score of -1.625 corresponds to approximately the 5.2nd percentile, meaning that a pregnancy lasting 240 days is shorter than 94.8% of all pregnancies.
04

Convert Days to Z-scores for Part (b)

To find the percentage of pregnancies lasting between 240 and 270 days, we convert both values to z-scores. For \(x = 240\), z-score is already calculated as -1.625. For \(x = 270\), \(z = \frac{270 - 266}{16} = \frac{4}{16} = 0.25\).
05

Calculate the Area Between Z-scores for Part (b)

Using a standard normal distribution table or calculator, we find the area to the left of \(z = 0.25\) which is approximately 0.5987, and the area to the left of \(z = -1.625\) which is approximately 0.052. Subtract these values: 0.5987 - 0.052 = 0.5467. Thus, approximately 54.67% of pregnancies last between 240 and 270 days.
06

Determine the Z-score for 80th Percentile for Part (c)

For finding the length of the longest 20% of pregnancies, look up the z-score that corresponds to the 80th percentile (since 100% - 20% = 80%). From z-tables or calculators, the z-score for the 80th percentile is about 0.8416.
07

Convert the Z-score Back to Days for Part (c)

Convert the z-score of 0.8416 to days using the formula: \[ x = z \times \sigma + \mu \]. So, \(x = 0.8416 \times 16 + 266 = 13.4656 + 266 \approx 279.47\). Rounded to the nearest day, the longest 20% of pregnancies last 280 days or more.

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Key Concepts

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

z-score
Understanding the concept of the z-score is fundamental in statistics, especially when working with normal distribution problems. A z-score measures how many standard deviations an element is from the mean. It's a way of standardizing scores on the same scale.
To calculate the z-score, use the formula:
  • \( z = \frac{x - \mu}{\sigma} \)
where:
  • \( x \) is the value being compared,
  • \( \mu \) is the population mean, and
  • \( \sigma \) is the population standard deviation.
In the context of pregnancy length, if you want to know how unusual a 240-day pregnancy is, you calculate its z-score. A negative z-score, like -1.625 for a 240-day pregnancy, informs us that the duration is shorter than the mean.
percentile rank
Percentile rank provides insight into how a particular score compares with others in a data set. It represents the percentage of scores that fall below a specific value in a dataset. For example, in our exercise, a pregnancy of 240 days is at about the 5.2nd percentile, meaning only 5.2% of pregnancies are shorter. The percentile rank is obtained by converting the z-score to this percentage. You can achieve this using a standard normal distribution table or statistical software. Understanding percentiles helps in assessing where a particular score stands in a distribution and can be better than using raw scores.
mean and standard deviation
The mean and standard deviation are central to understanding any normal distribution. The mean (\(\mu\)) is the average of all values in a data set and provides a central point of reference. In natural occurrences, like the length of pregnancies, it represents the typical amount of days to expect from conception to birth.
The standard deviation (\(\sigma\)) indicates how spread out the values are around the mean.
  • A small standard deviation means the data points are close to the mean.
  • A larger standard deviation indicates more variability in the data.
In the normal distribution of pregnancy lengths, a mean of 266 days with a standard deviation of 16 days shows a fairly tight clustering around the mean. Most pregnancies will last between (mean 卤 1 standard deviation) 250 to 282 days.
probability calculations
Calculating probabilities for normal distributions involves determining the area under the curve for the intervals of interest. We've seen this with calculating the probability that a pregnancy lasts between specific days.
  • To find the probability between two values, convert both to z-scores.
  • Use a standard normal distribution table to find the probabilities for each z-score.
  • Finally, subtract the smaller probability from the larger to get the area (probability) between them.
For example, the probability that a pregnancy lasts between 240 and 270 days was calculated by finding the area between their respective z-scores. The area provides the proportion of the population or the likelihood of a specific range of values, translated into a percentage for intuitive understanding. This allows for informed predictions based on statistical data.

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Most popular questions from this chapter

Brush your teeth The amount of time Ricardo spends brushing his teeth follows a Normal distribu- - tion with unknown mean and standard deviation. Ricardo spends less than one minute brushing his teeth about 40\(\%\) of the time. He spends more than two minutes brushing his teeth 2\(\%\) of the time Use this information to determine the mean and standard deviation of this distribution.

I think I can! An important measure of the performance of a locomotive is its adhesion," which is the locomotive's pulling force as a multiple of its weight. The adhesion of one 4400 -horsepower diesel locomotive varies in actual use according to a Normal distribution with mean \(\mu=0.37\) and standard deviation \(\sigma=0.04 .\) For each part that follows, sketch and shade an appropriate Normal distribution. Then show your work. (a) For a certain small train鈥檚 daily route, the locomotive needs to have an adhesion of at least 0.30 for the train to arrive at its destination on time. On what proportion of days will this happen? Show your method. (b) An adhesion greater than 0.50 for the locomotive will result in a problem because the train will arrive too early at a switch point along the route. On what proportion of days will this happen? Show your method. (c) Compare your answers to (a) and (b). Does it make sense to try to make one of these values larger than the other? Why or why not?

According to the Los Angeles Times, speed limits on California highways are set at the 85th percentile of vehicle speeds on those stretches of road. Explain what that means to someone who knows little statistics.

Eleanor scores 680 on the SAT Mathematics test. The distribution of SAT scores is symmetric and single-peaked, with mean 500 and standard deviation 100. Gerald takes the American College Testing (ACT) Mathematics test and scores 27. ACT scores also follow a symmetric, single-peaked distribution鈥攂ut with mean 18 and standard deviation 6. Find the standardized scores for both students. Assuming that both tests measure the same kind of ability, who has the higher score?

Cool pool? Coach Ferguson uses a thermometer to measure the temperature (in degrees Celsius) at 20 different locations in the school swimming pool. An analysis of the data yields a mean of \(25^{\circ} \mathrm{C}\) and a standard deviation of \(2^{\circ} \mathrm{C}\) . Find the mean and standard deviation of the temperature readings in degrees Fahrenheit (recall that \(^{\circ} \mathrm{F}=(9 / 5)^{\circ} \mathrm{C}+32 )\)
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