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Suppose that X has the normal distribution for which the mean is 1 and the variance is 4. Find the value of each of the following probabilities:

a. Pr(X ≤ 3)
b. Pr(X > 1.5)
c. Pr(X = 1)
d. Pr(25)
e. Pr(X ≥ 0)
f. Pr(−10.5)
g. Pr(|X| ≤ 2)
h. Pr(1≤−2X + 3 ≤ 8)

User MrMalith
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Final answer:

To find the value of each probability, we can use the properties of the normal distribution. The normal distribution is a continuous probability distribution that is symmetrical and bell-shaped. We can calculate the z-scores for the given values and use them to find the corresponding probabilities using the standard normal distribution table or a calculator.

Step-by-step explanation:

To find the value of each probability, we can use the properties of the normal distribution. The normal distribution is a continuous probability distribution that is symmetrical and bell-shaped. First, let's calculate the z-scores for the given values.

  • a. Pr(X ≤ 3): To find this probability, we need to calculate the z-score for 3. The z-score formula is z = (x - mean) / standard deviation. Therefore, z = (3 - 1) / 2 = 1. Next, we can look up the probability corresponding to this z-score in the standard normal distribution table or use a calculator to find Pr(Z ≤ 1), which is approximately 0.8413.
  • b. Pr(X > 1.5): Similarly, we calculate the z-score for 1.5 using the z-score formula: z = (1.5 - 1) / 2 = 0.25. To find the probability corresponding to this z-score, we need to calculate Pr(Z > 0.25), which is approximately 0.5987.
  • c. Pr(X = 1): Since the normal distribution is continuous, the probability of getting an exact value for a continuous random variable is zero. Therefore, Pr(X = 1) = 0.
  • d. Pr(25): Since 25 is not within the range of the distribution, its probability is also zero, Pr(25) = 0.
  • e. Pr(X ≥ 0): To find this probability, we can calculate the z-score for 0 using the z-score formula: z = (0 - 1) / 2 = -0.5. Then, we need to calculate Pr(Z ≥ -0.5), which is approximately 0.6915.
  • f. Pr(−10.5): Similar to the previous case, −10.5 is not within the range of the distribution, so its probability is also zero, Pr(−10.5) = 0.
  • g. Pr(|X| ≤ 2): Here, we want to find the probability that the absolute value of X is less than or equal to 2. This can be rewritten as Pr(-2 ≤ X ≤ 2). We can calculate the z-scores for -2 and 2 using the z-score formula: z1 = (-2 - 1) / 2 = -1.5 and z2 = (2 - 1) / 2 = 0.5. Finally, we subtract the probability corresponding to -1.5 from the probability corresponding to 0.5: Pr(-2 ≤ X ≤ 2) = Pr(-1.5 ≤ Z ≤ 0.5) = Pr(Z ≤ 0.5) - Pr(Z ≤ -1.5) = 0.6915 - 0.0668 = 0.6247.
  • h. Pr(1 ≤ -2X + 3 ≤ 8): To find this probability, we need to transform the inequality into two separate inequalities. First, let's solve the lower bound inequality: 1 ≤ -2X + 3. Subtracting 3 from both sides gives -2 ≤ -2X, and dividing by -2 (note that we flip the inequality sign since we are dividing by a negative number) gives 1 ≥ X. Next, let's solve the upper bound inequality: -2X + 3 ≤ 8. Subtracting 3 from both sides gives -2X ≤ 5, and dividing by -2 (again, we flip the inequality sign) gives X ≥ -2. Now we have the inequalities 1 ≥ X and X ≥ -2, which can be combined to form -2 ≤ X ≤ 1. We can calculate the z-scores for -2 and 1 using the z-score formula: z1 = (-2 - 1) / 2 = -1.5 and z2 = (1 - 1) / 2 = 0. Finally, we can find the probability corresponding to the z-scores: Pr(-2 ≤ X ≤ 1) = Pr(-1.5 ≤ Z ≤ 0) = Pr(Z ≤ 0) - Pr(Z ≤ -1.5) = 0.5000 - 0.0668 = 0.4332.
User Pablo Abdelhay
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