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
μ = 70
σ2 = 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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