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Lengths of pregnancies of women are normally distributed with a mean of 266 days and a standard deviation of 16 days.

What percentage to the nearest hundredth of children are born from pregnancies lasting more than 274 days?

What percentage to the nearest hundredth of children are born from pregnancies lasting less than 246 days?

User Vitalii Zurian
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2 Answers

20 votes
20 votes

To find the percentage of children who are born from pregnancies lasting more than 274 days or less than 246 days, we need to use the z-score formula to convert the lengths of pregnancies to standard units and the standard normal distribution to find the percentage of children who fall within the given range of pregnancy lengths.

The z-score formula is:

z = (x - mu) / sigma

where z is the z-score, x is the raw score, mu is the mean, and sigma is the standard deviation.

In this case, we are given that mu = 266 days and sigma = 16 days. We can use this information to find the z-score for the lower and upper bounds of the pregnancy length range, 246 days and 274 days, respectively.

For the lower bound of the pregnancy length range, we have:

z = (246 - 266) / 16

= -20 / 16

= -1.25

For the upper bound of the pregnancy length range, we have:

z = (274 - 266) / 16

= 8 / 16

= 0.5

Now we can use the standard normal distribution to find the percentage of children who have a z-score between -1.25 and 0.5. The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1, so the percentage of children who fall within the given pregnancy length range is the same as the percentage of children who have a z-score within this range.

To find the percentage of children who have a z-score between -1.25 and 0.5, we can use a standard normal distribution table to look up the z-scores and find the corresponding probabilities. The standard normal distribution table shows the percentage of values that fall within a given range of z-scores, so to find the percentage of children who have a z-score between -1.25 and 0.5, we need to look up the percentage of values that fall within this range.

According to the standard normal distribution table, the percentage of values that fall within the range of z-scores -1.25 to 0.5 is approximately 0.38. Therefore, the percentage of children who are born from pregnancies lasting more than 274 days or less than 246 days is approximately 38%, to the nearest hundredth.

User Boedy
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13 votes
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To solve this problem, we need to use the z-score formula to convert the given values to standard units. This will allow us to use the normal distribution table to find the percentage of pregnancies that fall within a certain range of lengths.

First, let's find the z-score for a pregnancy lasting more than 274 days:

z = (274 - 266) / 16 = 0.5

Next, let's use the normal distribution table to find the percentage of pregnancies that are longer than 274 days. Since the z-score we calculated is positive, we need to look in the right tail of the distribution. From the table, we find that the percentage of pregnancies that are longer than 274 days is approximately 30.85%.

Now let's find the z-score for a pregnancy lasting less than 246 days:

z = (246 - 266) / 16 = -1.25

To find the percentage of pregnancies that are shorter than 246 days, we need to look in the left tail of the distribution. From the table, we find that the percentage of pregnancies that are shorter than 246 days is approximately 8.38%.

Therefore, the percentage of children that are born from pregnancies lasting more than 274 days is approximately 30.85%, and the percentage of children that are born from pregnancies lasting less than 246 days is approximately 8.38%.

User Ben Hare
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