A population of values has a normal distribution with μ = 8.2 and σ = 30.2 . You intend to draw a random sample of size n = 28 . Find the probability that a single randomly selected value is greater than -0.9. P(X > -0.9) = Find the probability that a sample of size n = 28 is randomly selected with a mean greater than -0.9. P(M > -0.9) = Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.
Solution :
Given that ,
mean = = 8.2
standard deviation = = 30.2
n = 28
M = 8.2 and
M = / n = 30.2 / 28 = 5.7073
P(M > -0.9) = 1 - P(M < -0.9)
= 1 - P(( - _{} ) / _{} < (-0.9 - 8.2) / 5.7073)
= 1 - P(z < -1.594)
= 1 - 0.0555
= 0.9445
P(M > -0.9) = 0.9445
Probability = 0.9445
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