The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. Out of 50 pregnancies, how many would you expect to last less than 250 days?
2 answers
Assume that the heights of women are normally distributed with a mean of 63.9 inches and a standard deviation of 2.5 inches. The U.S. Army requires that the heights of women be between 58 and 80 inches. If 200 women want to enlist in the U.S. Army, how many would you expect to meet the height requirements?
Use z-scores. For your first problem, determine z using the formula:
z = (x - mean)/sd
...where x = 250, mean = 268, and sd = 15.
Once you have z, check the z-table for the probability (remember the first problem is asking less than 250, so keep that in mind when checking the table). Once you have the probability, multiply 50 by that value you find to get the number you need.
The second problem is similar to the first, with the exception of finding two z-scores instead of one.
Find z using 58 for x and also find z using 80 for x. Then use the z-table to find the probability between the two z-scores. Once you have the probability, multiply 200 by that value you find to get the number you need.
I hope this will help get you started.
z = (x - mean)/sd
...where x = 250, mean = 268, and sd = 15.
Once you have z, check the z-table for the probability (remember the first problem is asking less than 250, so keep that in mind when checking the table). Once you have the probability, multiply 50 by that value you find to get the number you need.
The second problem is similar to the first, with the exception of finding two z-scores instead of one.
Find z using 58 for x and also find z using 80 for x. Then use the z-table to find the probability between the two z-scores. Once you have the probability, multiply 200 by that value you find to get the number you need.
I hope this will help get you started.
1 answer