Question

the lengths of a particular animals pregnancies are approximately normally distributed, with mean $mu = 261$ days and standard deviation $sigma = 16$ days.
(a) what proportion of pregnancies lasts more than 281 days?
(b) what proportion of pregnancies lasts between 253 and 269 days?
(c) what is the probability that a randomly selected pregnancy lasts no more than 241 days?
(d) a \very preterm\ baby is one whose gestation period is less than 221 days. are very preterm babies unusual?
(a) the proportion of pregnancies that last more than 281 days is 0.1056.
(round to four decimal places as needed.)
(b) the proportion of pregnancies that last between 253 and 269 days is 0.3830.
(round to four decimal places as needed.)
(c) the probability that a randomly selected pregnancy lasts no more than 241 days is 0.1056.
(round to four decimal places as needed.)
(d) the probability of a \very preterm\ baby is 0.0062. this event
(round to four decimal places as needed.)

Explanation:

Step1: Determine the z - score formula

The z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(x\) is the value from the data set, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step2: Calculate the z - score for \(x = 221\)

Given \(\mu=261\) and \(\sigma = 16\), then \(z=\frac{221 - 261}{16}=\frac{-40}{16}=- 2.5\)

Step3: Find the probability using the standard normal distribution table

We want to find \(P(X\lt221)\), which is equivalent to \(P(Z\lt - 2.5)\)
Looking up \(z=-2.5\) in the standard - normal distribution table (or using a calculator with a normal - distribution function, e.g., \(P(Z\lt z)\) for \(z=-2.5\)), we get \(P(Z\lt - 2.5)=0.0062\)

Step4: Determine if the event is unusual

An event is considered unusual if its probability \(P\lt0.05\)

Answer:

The probability of a "very preterm" baby is \(0.0062\). This event would be unusual because the probability is less than \(0.05\).