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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%
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