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3.3-5 If X is normally distributed with a mean of 6 and a variance of 25 , find (a) P(6≤X≤12). (b) P(0≤X≤8). (c) P(−221). (e) P(∣X−6∣<5). (f) P(∣X−6∣<10). (g) P(∣X−6∣<15) (h) P(∣X−6∣<12.41).

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Final answer:

To find the probabilities, we need to standardize the values of X using the mean and standard deviation given. Then, we can use the standard normal distribution table to find the probabilities.

Step-by-step explanation:

To find the probabilities in this problem, we need to standardize the values of X using the mean and standard deviation given. Let's call the standardized variable Z. To find P(6≤X≤12), we need to find P(0≤Z≤1.2). From the standard normal distribution table, we find that the z-score corresponding to a value of 1.2 is approximately 0.8849. Therefore, P(0≤Z≤1.2) = 0.8849 - 0.5000 = 0.3849.

To find P(0≤X≤8), we need to find P(-2.4≤Z≤0.4). From the standard normal distribution table, we find that the z-score corresponding to a value of 0.4 is approximately 0.6554, and the z-score corresponding to a value of -2.4 is approximately 0.0082. Therefore, P(-2.4≤Z≤0.4) = 0.6554 - 0.0082 = 0.6472.

For P(|X-6|<5), we can rewrite it as P(-5

For P(|X-6|<10), we can rewrite it as P(-10

Similarly, for P(|X-6|<15), we can rewrite it as P(-15

For P(|X-6|<12.41), we can rewrite it as P(-12.41

User Evengard
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7 votes

Final answer:

To find the probabilities in a normal distribution with mean 6 and variance 25, we need to standardize the values using the z-score formula. With the z-scores, we can calculate the probabilities for each question.

Step-by-step explanation:

To solve this problem, we need to standardize the normal distribution using the formula z = (x - mean) / standard deviation. Let's calculate the values for each part of the question:

(a) P(6≤X≤12) = P((6-6)/5≤Z≤(12-6)/5) = P(0≤Z≤1.2)

(b) P(0≤X≤8) = P((0-6)/5≤Z≤(8-6)/5) = P(-1.2≤Z≤0.4)

(c) P(−221) = 0 since there is no value for X less than the lower bound (mean - 3 standard deviations)

(e) P(|X-6|<5) = P(1≤Z≤2) + P(-2≤Z≤-1) = P(1≤Z≤2) + P(1≤Z≤-2)

(f) P(|X-6|<10) = P(-1≤Z≤3)

(g) P(|X-6|<15) = P(-3≤Z≤5)

(h) P(|X-6|<12.41) = P(-0.7≤Z≤2.48)

User Krejko
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