User
Let X be a random variable normally distributed with mean 1 and variance 2 . What is the probability that X belongs between 0.5 and 2?
To find the probability that X belongs between 0.5 and 2, we need to standardize the range of values using the z-score.
The formula for the z-score is:
z = (X - μ) / σ
Where:
X is the value of the random variable
μ is the mean
σ is the standard deviation (square root of the variance)
In this case, the mean (μ) is 1 and the standard deviation (σ) is the square root of 2, which is approximately 1.414.
Let's calculate the z-scores for 0.5 and 2:
z1 = (0.5 - 1) / 1.414 = -0.3536
z2 = (2 - 1) / 1.414 = 0.7071
Now, we need to find the probabilities associated with these z-scores using a standard normal distribution table or a calculator.
The probability that X belongs between 0.5 and 2 can be calculated as the difference between the cumulative probabilities at z2 and z1:
P(0.5 < X < 2) = P(0 < Z < 0.7071) - P(0 < Z < -0.3536)
Using a standard normal distribution table or calculator, we find:
P(0 < Z < 0.7071) ≈ 0.259
P(0 < Z < -0.3536) ≈ 0.363
Therefore, the probability that X belongs between 0.5 and 2 is approximately:
P(0.5 < X < 2) ≈ 0.259 - 0.363 ≈ -0.104.
Note that this probability cannot be negative, so there may be a mistake in the calculations or assumptions made. Please double-check the input values and the steps followed to verify the correct solution.
To find the probability that X belongs between 0.5 and 2, we need to standardize the random variable X and use the standard normal distribution.
Let's denote the standardized random variable as Z. The standardization formula is given by:
Z = (X - μ) / σ
Where μ is the mean of X and σ is the standard deviation of X.
For the given problem, we have μ = 1 (mean) and σ^2 = 2 (variance).
Taking the square root of the variance, we find that σ = sqrt(2) ≈ 1.414.
Now, let's calculate the standardized value for Z when X = 0.5 and when X = 2.
Z1 = (0.5 - 1) / 1.414 ≈ -0.3536
Z2 = (2 - 1) / 1.414 ≈ 0.7071
We can now look up the corresponding probabilities for Z1 and Z2 in the standard normal distribution table (also known as the Z-table).
Looking up the Z-table, we find that the probability corresponding to Z = -0.3536 is approximately 0.3632, and the probability corresponding to Z = 0.7071 is approximately 0.7602.
To find the probability that X belongs between 0.5 and 2, we subtract the probability corresponding to Z1 from the probability corresponding to Z2:
P(0.5 ≤ X ≤ 2) = P(Z1 ≤ Z ≤ Z2)
= P(Z ≤ Z2) - P(Z ≤ Z1)
= 0.7602 - 0.3632
≈ 0.397
Therefore, the probability that X belongs between 0.5 and 2 is approximately 0.397, or 39.7%.