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A normal random variable X has an unknown mean and standard deviation. The probability that X exceeds 3 is 0.9772, and the probability that X exceeds 6 is 0.9332. Find the mean and standard deviation.

User Asadnwfp
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By analyzing the given probabilities and utilizing properties of the normal distribution, we found the mean of X to be around 2.0 and the standard deviation to be roughly 0.707.

Convert probabilities to cumulative distribution function (CDF) values:

The given probabilities represent the area to the right of 3 and 6 under the standard normal curve. Therefore, they correspond to CDF values:

P(X > 3) = 0.9772 = CDF(3, μ, σ)

P(X > 6) = 0.9332 = CDF(6, μ, σ)

where μ and σ are the unknown mean and standard deviation, respectively.

Solve for mean and standard deviation using properties of the normal distribution:

We can exploit the symmetry of the normal distribution and the relationship between CDF values and standard normal scores (z-scores). Since the area between 3 and 6 is smaller than the total area to the right of 3, the mean (μ) must lie between 3 and 6. Additionally, the z-scores corresponding to the given CDF values can be obtained using a standard normal table or numerical methods.

Let z_3 and z_6 be the z-scores for P(X > 3) and P(X > 6), respectively. Then:

z_3 ≈ 2.5 (corresponding to 0.9772 in the standard normal table)

z_6 ≈ 1.5 (corresponding to 0.9332)

Using the relationship between z-scores, mean, and standard deviation:

μ = (z_6 + z_3) / 2

σ = (|z_3| - |z_6|) / sqrt(2)

Calculate the values:

μ = (1.5 + 2.5) / 2 ≈ 2.0

σ = (2.5 - 1.5) / sqrt(2) ≈ 0.707

Therefore, the mean (μ) of the normal random variable X is approximately 2.0 and the standard deviation (σ) is approximately 0.707.

User Steve Grove
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