Find solutions to your problems with the help of IDNLearn.com's knowledgeable users. Ask any question and receive comprehensive, well-informed responses from our dedicated team of experts.
Sagot :
Answer:
So, the probability that the total length of the 9 trials is at least 225 days is approximately 0.0436, or 4.36%
Step-by-step explanation:
To solve this problem, we need to find the probability that the total length of 9 trials is at least 225 days, given that each trial has a mean duration of 21 days and a standard deviation of 7 days.
First, let's denote:
- \( X_i \) as the duration of the \( i \)-th trial.
- The total duration of 9 trials as \( S = X_1 + X_2 + \cdots + X_9 \).
Since each \( X_i \) follows a normal distribution with mean \( \mu = 21 \) days and standard deviation \( \sigma = 7 \) days, the sum \( S \) of these 9 trials will follow a normal distribution with:
- Mean of \( 9 \times 21 = 189 \) days (since \( \mu_S = n \mu = 9 \times 21 \)).
- Standard deviation of \( \sigma_S = \sqrt{n} \times \sigma = \sqrt{9} \times 7 = 21 \) days.
Now, we need to find \( P(S \geq 225) \).
Calculate the z-score for \( S = 225 \):
\[ z = \frac{225 - 189}{21} = \frac{36}{21} \approx 1.7143 \]
Using the standard normal distribution table or a calculator, find the probability corresponding to \( z = 1.7143 \):
\[ P(S \geq 225) = 1 - P(Z < 1.7143) \]
From the standard normal distribution table:
\[ P(Z < 1.7143) \approx 0.9564 \]
Therefore,
\[ P(S \geq 225) = 1 - 0.9564 = 0.0436 \]
So, the probability that the total length of the 9 trials is at least 225 days is approximately 0.0436, or 4.36%
We appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Thanks for visiting IDNLearn.com. We’re dedicated to providing clear answers, so visit us again for more helpful information.
