Question

The weight of a 5th grader is normally distributed with a mean of 69 pounds and a variance of 71 pounds^2. Let weight, in pounds, be represented by random variable X.

P(65 < X < x2) = 0.0990. Find x2 (in pounds).

Answer 1

Solution:- Given that information mean µ = 69, variance = 71,

=> σ = sqrt(71) - 8.4261

=> P(65 < X < X2) = 0.0990

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converting the scores in to z-scores

P[(65-69)/8.4261 < z < (x2-69)/8.4261] = 0.0990

P(-0.4747 < z < (x2-69)/8.4261) = 0.0990

P(z < (x2-69)/8.4261) - P(z < -0.4747) =0.0990

P(z < (x2-69)/8.4261) - 0.3192 = 0.0990

P(z < (x2-69)/8.4261) = 0.0990 + 0.3192

P(z < (x2-69)/8.4261) = 0.4182

(x2-69)/8.4261 = invNorm(0.4182)

(x2-69)/8.4261 = -0.2065

x2 = 69 - (0.2065*8.4261) = 67.2600