Question

The lengths of pregnancies in a small rural village are normally distributed with a mean of 266 days and a standard deviation of 16 days. Let X be the length of a randomly recorded pregnancy in the village.

a.) If a pregnancy randomly chosen in the village, find the probability that it lasted beyond 290 days.

b.) If a pregnancy randomly chosen in the village, find the probability that it lasted between 258 and 266 days.

c.) The 71st percentile pregnancy length in this village is

Answer 1

The question involves calculating probabilities and percentiles for a normally distributed variable, which in this case, is the length of pregnancies in a village. The calculations involve finding Z-scores and using them to determine probabilities and percentile values.

Step-by-step explanation:

The lengths of pregnancies in a small rural village are normally distributed with a mean of 266 days and a standard deviation of 16 days. Let X be the length of a randomly recorded pregnancy in the village.

1. 
   
3. To find the probability that a pregnancy lasted beyond 290 days, we calculate the Z-score for 290 days using Z = (X - μ) / σ (where μ is the mean and σ is the standard deviation), then use the Z-score to find the corresponding probability from the standard normal distribution table or a calculator's distribution function.
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6. To find the probability that a pregnancy lasted between 258 and 266 days, we calculate the Z-scores for both values and find the probability for each. The probability that X is between 258 and 266 is the difference between the two probabilities.
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9. The 71st percentile pregnancy length is the value of X that corresponds to the 71st percentile on the standard normal distribution. This can be found using the Z-score associated with the 71st percentile and the given mean and standard deviation.