how would i do about this problem can someone explain it to me or show me because i keep getting weird answers with an e at the end. It makes no sence.
This question is: The length of human pregnancies from conception to birth varies according to a distrubition that is approximately normal with mean 266 days and standard deviation 16 days. Call the length of a randomly chosen pregnancy Y. What is P(Y>300)?
4
how did you get 4 ?
To solve this problem, we need to use the concept of the normal distribution and standard deviation.
The mean of the normal distribution is given as 266 days, and the standard deviation is 16 days. This means that the distribution of pregnancy lengths is centered around 266 days, with 68% of pregnancies falling within one standard deviation of the mean (between 250 and 282 days), and 95% of pregnancies falling within two standard deviations of the mean (between 234 and 298 days).
To find the probability that a randomly chosen pregnancy lasts longer than 300 days (P(Y > 300)), we need to determine how many standard deviations away from the mean 300 days is. We can do this by calculating the z-score.
The z-score formula is: z = (X - μ) / σ
Where:
X is the value we are interested in (in this case, 300 days)
μ is the mean of the distribution (266 days)
σ is the standard deviation of the distribution (16 days)
Plugging in the values, we get:
z = (300 - 266) / 16
z = 34 / 16
z ≈ 2.125
Now we need to find the probability associated with this z-score. We can look up this probability in a standard normal distribution table or use a calculator/ software.
Looking up the z-score of 2.125 in the table or using a calculator, you will find that the probability associated with this z-score is approximately 0.9838.
So, P(Y > 300) is approximately 0.9838 or 98.38%.
Therefore, the answer to the question is approximately 0.9838 or 98.38%.