Durations of Pregnancies The lengths of pregnancies are normal...

Question

Durations of Pregnancies The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days.

a. In a letter to “Dear Abby,” a wife claimed to have given birth 308 days after a brief visit from her husband, who was working in another country. Find the probability of a pregnancy lasting 308 days or longer. What does the result suggest?

Accepted Answer

Hello there. Today we're gonna solve the following practice problem together. So, first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. The average lifespan of a certain species of trees in a forest is normally distributed with a mean of 50 years and a standard deviation of 8 years. A researcher claims to have found a tree in this forest that is 72 years old. Find the probability of finding a tree of this age or older in this forest. Awesome. So it appears for this particular problem we're ultimately trying to figure out how to find the probability of finding a tree. So we're trying to find the probability of finding a tree that is 72 years old or older in this specific forest. So what is the probability of finding a tree within the same forest that is 72 years old or older? So now that we know that we're ultimately trying to solve for this probability value, let's read off our multiple choice answers to see what our final answer might be. A is 0.0016. B is 0.0022, C is 0.0030, and finally D is 0.0040. Awesome. So first off, we need to standardize the value 72 to a Zco value. So as we should recall, the Z2 equation states that Z is equal to X minus m divided by sigma. Awesome, which when we plug in our known variables, this will be equal to 72, because that's our value for X minus 50, which 50 is our mean. So mu denotes the mean. X denotes our value that we are told by the problem itself, so it claims that they found a tree in the forest that is 72 years old, and sigma, of course, is the standard deviation, so we need to take 72 minus 50 divided by 8. And when we plug that into our calculator, we will get 2.75 when we round to two decimal places. So now we need to recall and note that the probability we want is the probability P of X is greater than or equal to 72 is equal to the probability of Z is greater than or equal to 2.75. Fantastic. So now we need to use the table for positive Z scores. So referring to this particular table, we will find that when Z is 2.7 and referring to 0.05, we will have an answer of 0.9970. So that means that using our table, we can go ahead and write that the probability of Z. is less than 2.75 is, so once again, the probability. Of Z is less than 2.75 will be equal to 0.9970, and this is a value we pull from our table. So with that said, Therefore, we go ahead and write that P of Z is greater than or equal to 2.75 will be equal to 1 minus 0.9970. So that will mean that P of Z is greater than or equal to 2.75 will be equal to 0.0030. And that is our final answer. Hooray, we did it. So therefore, the probability of finding a tree that is age 72 years or older is approximately 0.0030, and this corresponds to multiple choice answer letter C, which once again is 0.0030. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

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Key Concepts

Normal Distribution

Z-Score

Probability

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