Question

Let \(x\) denote the duration of a randomly selected pregnancy (the time elapsed between conception and birth). Accepted values for the mean value and standard deviation of \(x\) are 266 days and 16 days, respectively. Suppose that the probability distribution of \(x\) is (approximately) normal. a. What is the probability that the duration of a randomly selected pregnancy is between 250 and 300 days? b. What is the probability that the duration is at most 240 days? c. What is the probability that the duration is within 16 days of the mean duration? d. A "Dear Abby" column dated January 20, 1973, contained a letter from a woman who stated that the duration of her pregnancy was exactly 310 days. (She wrote that the last visit with her husband, who was in the navy, occurred 310 days before the birth of her child.) What is the probability that the duration of pregnancy is at least 310 days? Does this probability make you a bit skeptical of the claim? e. Some insurance companies will pay the medical expenses associated with childbirth only if the insurance has been in effect for more than 9 months ( 275 days). This restriction is designed to ensure that benefits are only paid if conception occurred during coverage. Suppose that conception occurred 2 weeks after coverage began. What is the probability that the insurance company will refuse to pay benefits because of the 275 -day requirement?

Short answer

The computed probabilities for a-e would be the viable answers here, but these values need to be looked up in a standard normal distribution table or calculated with a statistical software. Also, remember to always take the symmetry properties of the normal distribution into account and that probabilities always lie between 0 and 1.

Step-by-step solution

Translate into Z-scores

For each question part a-e, translate the given conditions into Z-scores by using the formula Z = (X - µ) / σ, where X is the duration, µ is the mean and σ is the standard deviation. Then, look up these Z-scores in the standard normal distribution table to get the probabilities.

Calculation for part (a)

P(250 < x < 300) = P(Z < (300-266)/16) - P(Z < (250-266)/16). Look up these Z-scores in the standard normal distribution table to find the probabilities.

Calculation for part (b)

P(x < 240) = P(Z < (240-266)/16). Look up this Z-score in the standard normal distribution table to find the probability.

Calculation for part (c)

P(250 < x < 282) = P(Z < (282-266)/16) - P(Z < (250-266)/16). Look up these Z-scores in the standard normal distribution table to find the probabilities.

Calculation for part (d)

P(x > 310) = 1 - P(Z < (310-266)/16). Look up this Z-score in the standard normal distribution table to find the corresponding probability and subtract it from 1.

Calculation for part (e)

P(x > 275) = 1 - P(Z < (275-266)/16). Look up this Z-score in the standard normal distribution table to find the corresponding probability and subtract it from 1.

Key concepts

Normal Distribution

When we talk about normal distribution in statistics, we're referring to a probability distribution that is symmetrical and bell-shaped. This distribution is centered around the mean, and it describes how the values of a variable are spread out. A key property of the normal distribution is that about 68% of the values fall within one standard deviation from the mean, 95% within two standard deviations, and 99.7% within three.

For example, in the context of pregnancy durations, if we assume that these durations are normally distributed with a mean of 266 days and a standard deviation of 16 days, most pregnancies will last between 250 (one standard deviation below the mean) and 282 days (one standard deviation above the mean). However, it's important to note that while the normal distribution is a common model, not all data fits this pattern perfectly. In the case of very extreme values, like a pregnancy lasting 310 days, they occur with much less frequency, as suggested by the 'tails' of the distribution.

Z-scores

A Z-score, also known as a standard score, is a way to quantify how many standard deviations a given value is from the mean. It's a universal measure that allows statisticians to compare data points from different normal distributions, or to assess where a single data point lies within its distribution.

In our pregnancy example, you would compute a Z-score for each duration to understand how unusual or common that duration is relative to the typical pregnancy duration (the mean). The Z-score is calculated using the formula: \( Z = (X - \mu) / \sigma \), where \( X \) is the value of interest, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. Positive Z-scores indicate values above the mean, while negative scores indicate values below it. The higher the absolute value of a Z-score, the more unusual the data point is. For instance, a Z-score of 2 or -2 would be quite rare as it lies far from the mean.

Mean and Standard Deviation

The mean is the arithmetic average of a set of values, and the standard deviation measures the amount of variation or dispersion from the mean. Together, these two statistics form the backbone of the normal distribution and are critical in calculating the Z-scores mentioned earlier.

For our pregnancy duration scenario, the mean is 266 days, suggesting that on average, pregnancies last this length. The standard deviation is 16 days, which means most pregnancies vary by 16 days shorter or longer than the mean. These two statistics provide a frame of reference for any given pregnancy duration. If we hear of a pregnancy lasting 310 days, we can use the mean and standard deviation to determine how far from 'average' this value is and calculate the associated probability. Both statistics also play a pivotal role in various probability calculations, as they allow us to quantify the likelihood of various ranges of outcomes.