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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. (a) At what percentile is a pregnancy that lasts 240 days (that's 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) 5.24th percentile, (b) 54.63%, (c) about 279.44 days.

Step by step solution

01

Understand the Distribution

We are given that the length of pregnancies follows a normal distribution with a mean (\(\mu\)) of 266 days and a standard deviation (\(\sigma\)) of 16 days. This information will help us calculate probabilities and percentiles.
02

Compute Z-score for 240 days (Part a)

To find the percentile for a 240-day pregnancy, we first compute the Z-score using the formula: \[ Z = \frac{X - \mu}{\sigma} \] where \(X = 240\), \(\mu = 266\), and \(\sigma = 16\). Plugging in the values, we get: \[ Z = \frac{240 - 266}{16} = \frac{-26}{16} = -1.625 \]
03

Determine Percentile from Z-score (Part a)

Using a standard normal distribution table or a calculator, we find the percentile for \( Z = -1.625 \). This corresponds approximately to the 5.24th percentile, meaning a 240-day pregnancy is at the 5.24th percentile.
04

Compute Z-scores for 240 and 270 days (Part b)

Next, we find the Z-scores for both 240 and 270 days. We already have that for 240 days: \(-1.625\). For 270 days: \[ Z = \frac{270 - 266}{16} = \frac{4}{16} = 0.25 \]
05

Calculate Probability Between Two Z-scores (Part b)

Using the Z-scores \(-1.625\) and \(0.25\), we find the area between these two points. From standard normal distribution tables: - The cumulative probability for \(Z = -1.625\) is approximately 0.0524. - The cumulative probability for \(Z = 0.25\) is approximately 0.5987. Thus, the probability between 240 and 270 days is \(0.5987 - 0.0524 = 0.5463\), or 54.63%.
06

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

To find how long the longest 20% of pregnancies last, we need the 80th percentile (since the longest 20% correspond to those above the 80th percentile). From a standard normal distribution table, a Z-score of approximately 0.84 corresponds to the 80th percentile.
07

Calculate the Length for Longest 20% (Part c)

Use the Z-score to find the actual length: \[ X = Z \times \sigma + \mu \] Plugging in the values: \[ X = 0.84 \times 16 + 266 = 13.44 + 266 = 279.44 \] So, the longest 20% of pregnancies last about 279.44 days.

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Key Concepts

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

Percentiles
Percentiles help us understand the position of a value within a distribution. In the context of human pregnancies, the percentile tells us how a specific pregnancy length compares with others.
For example, if a pregnancy lasts 240 days and we determine it is at the 5.24th percentile, this means that only about 5.24% of pregnancies are as short or shorter than this one.
  • Percentile rank shows relative standing.
  • Helps compare different data points in a distribution.
Percentiles are especially useful in normally distributed data because they make comparisons simpler. Understanding percentiles can help when interpreting standardized test scores or medical measurements, as you can quickly grasp how common or rare a certain measurement is. In the given exercise, when asked about the longest 20% of pregnancies, we are essentially looking at the 80th percentile and above.
Z-score
A Z-score is a statistical measurement that describes a value's relation to the mean of a group of values. It tells us how many standard deviations a value is from the average. In our example, the Z-score for a 240-day pregnancy is \( Z = \frac{-26}{16} = -1.625 \).
  • The formula: \( Z = \frac{X - \mu}{\sigma} \)
  • A negative Z-score indicates the value is below the mean.
  • A positive Z-score indicates it is above the mean.
Computing Z-scores is crucial for determining percentiles because it allows us to convert a raw score into a format that can be compared against the standard normal distribution. Z-scores are foundational in tasks like calculating probabilities and interpreting results within the context of the entire dataset.
Standard Deviation
Standard deviation plays a key role in understanding how much variation or "spread" there is from the average (mean). In simpler terms, it tells us how much the data points in a distribution are clustered around the mean or spread out.
In the exercise, the standard deviation of 16 days shows us that most pregnancies fall within 16 days either side of the mean (266 days).
  • Measures variability within a set of data.
  • Provides context for assessing Z-score results.
  • A small standard deviation indicates data points are close to the mean.
  • A large standard deviation indicates data is spread out.
When dealing with a normal distribution, such as pregnancy lengths, the standard deviation is a handy parameter to gauge the likelihood of certain lengths occurring. For instance, about 68% of data in a normal distribution lies within one standard deviation from the mean, and this rule helps in making quick predictions about data behavior under normal conditions.
Cumulative Probability
Cumulative probability refers to the probability that a random variable will take a value less than or equal to a certain amount. Using the normal distribution, we can find the cumulative probability for a specific Z-score.
In the context of the example, cumulative probabilities are used to measure how much of the pregnancy lengths fall between two Z-scores (such as between 240 and 270 days).
  • Shows the likelihood of a value falling within a particular range.
  • Integral to calculating percentiles and probabilities.
  • Helps visualize the spread and concentration of probabilities.
For instance, if the Z-score for 240 days has a cumulative probability of approximately 0.0524, and for 270 days it is about 0.5987, we find the probability of pregnancies lasting between these two intervals by subtracting the smaller cumulative probability from the larger one, yielding a result of 54.63%. This process highlights the importance of cumulative probability in understanding the probability distribution of a normally distributed variable.

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