What proportion of pregnancies last beyond 223 days? P(X > 223 days) =

Question

The lengths of pregnancies in a small rural village are normally distributed with a mean of 266 days and a standard deviation of 15 days. A distribution of values is normal with a mean of 266 and a standard deviation of 15.

What proportion of pregnancies last beyond 223 days? 

P(X > 223 days) = 

Answer 1

Hello Abby,

While reading over the problem, I always like to make note of any key words or values that we are given. In this case, "normal distribution," caught my attention, as well as the values for the mean and standard deviation (266 and 15 respectively). So, what this tells me is that we can use the normal distribution and its properties to answer the question.

There are different ways to approach the problem and I will explain two ways.

My first approach will be under the assumption that I do have access to a TI-84 Calculator.

1. Determine which calculator function to use.
2. In this case, we are dealing with a normal distribution, and we are given both the mean and standard deviation (st.dev). So we can determine that we will use one of the two "normal..." functions.
3. To determine if we will use "normalpdf(..)" and "normalcdf(..)" we will then look at the probability statement. The probability statement which we are given, P ( X > 223 days) uses the ">" (greater than inequality). This means that the variable X can be a cumulation of values, and not just one single value. So this ultimately tells me that we will use the "normalcdf(...)" function which just means normal cumulative distribution function.
4. Determine which buttons to use to get to the calculator function.
5. Press "on" to turn on your calculator.
6. Press the "2nd" button (which on my calculator is colored blue).
7. Press the "vars" button (which should have a blue "distr" label just above the button).
8. Scroll down and hit the "enter" button on "2:normalcdf(" function.
9. Determine what values to input.
10. For the "lower" section, the value that we input here should be the lower limit value. This is essentially is a fancy way of saying the starting point value (so which value is going to my minimal value?).
11. Considering that our probability statement is: P( X > 223), this tells me that the value 223 will be the starting point of the area/probability that I am trying to find.
12. We input 223.
13. For the "upper" section, the value that we input here should be the upper limit value.
14. So what is the highest possible value that is going to be included in this distribution? Well, a normal distribution essentially goes from (-) negative infinity to (+) positive infinity. And we are interested in the values that are greater than 223, so any value bigger than 223. This means we are interested in (+) positive infinity. How do we denote that in the calculator?
15. I like to use 1E99 to indicate as infinity, which in scientific notation is 1 * 10^99 (or 1 times 10 to the 99th power). In the calculator press, the "1" button --> "2nd" (blue button) --> "," (comma button with a blue EE labeled on top) --> "9" --> "9".
16. We input 1E99.
17. We then input the mean and standard deviation values as "266" and "15".
18. Scroll down and enter the "paste" function and we should get an answer of "0.9979258365."
19. Based on rounding rules for probability (which is four decimal places), our final answer should be 0.9979.

My second approach will be under the assumption that I do not have access to a TI-84 Calculator.

1. As a visual learner, I first like to draw out a bell-shaped curve when dealing with a normal distribution problem.
2. So I draw out the curve with the mean of 266 labeled in the middle. Then I label the standard deviations. I go as far out as 3 standard deviations below and above the mean. So the values (including the standard deviations and mean) from left to right should read: 221, 236, 251, 266, 281, 296, 311. (Note: 1 standard deviation represents 15 days).
3. Next I shade in the region that I am interested in. In this case, it will be the area to the right of the value "223" since my probability statement reads: P ( X > 223). This just gives me a better visual of what area/probability should represent.
4. Moving forward. I will use the z-score formula to help me find the probability.
5. Reminder: The z-score formula is x-μ/ σ.
6. We substitute the variables with the values given from the problem: 223 - 266 / 15. We should have a z-score of -2.87 (after following rounding rules).
7. Then we will utilize a z-score table/chart. If you do not have one available, look up "z-score table" online and you should have access to free ones.
8. We will look under the "Negative Z-Score Table" because we do in fact have a (-) negative z-score. Following the first column, find the values under the z symbol for " -2.8" and under the first row find the value "0.07." At the intersection of these two, you should find a value of ".00205."
9. Now because we are interested in the probability (or shaded area to the right), we are not done. Because if we look closely at the graph illustrated from the "Negative Z-Score Table," the shaded area is to the left of the X value NOT the right. So we will then subtract 1 - 0.00205 because (the total area of a probability distribution should be 1).
10. The remaining area should be "0.99795" and following rounding rules (of four decimal values), we should have a final answer of 0.9980.