Question

Let X be a random variable with a mean distribution of mean μ =
70 and variance σ2 = 15.

d) Imagine a symmetric interval around the mean (μ ± c) of the
distribution described above. Find the value of c such that the
probability is about 0.2 that X is in this interval.

Please explain how to get the answer

Answer #1

Solution:

μ = 70

σ^{2} = 15

σ = 15

P(μ - c < X < μ + c ) = 0.2

Since , the distribution is symmetric about mean , we can write,

P(X < μ - c ) = 0.4 and P(X > μ - c ) = 0.4

Consider , P(X < μ - c ) = 0.4

P(X - μ < -c ) = 0.4

P([(X - μ)/σ < -c/σ ) = 0.4

P(Z < -c/σ ) = 0.4

Use z table . Search where the probability 0.4 . See corresponding z value .

P(Z < -0.25) = 0.4

Comparing , we get

-c/σ = -0.25

c/σ = 0.25

c = 0.25 * σ = 0.25 * 15 = 0.96824583655 = 0.97

**c = 0.97**

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