A population of values has a normal distribution with μ = 101.4 and σ = 82.4 . You intend to draw a random sample of size n = 129 .
Find the probability that a single randomly selected value is greater than 96.3. P(X > 96.3) =
Find the probability that a sample of size n = 129 is randomly selected with a mean greater than 96.3. P( ¯ x > 96.3)=
Enter your answers as numbers accurate to 4 decimal places.
Solution :
Given that ,
mean = = 101.4
standard deviation = = 82.4
P(x >96.3 ) = 1 - P(x <96.3 )
= 1 - P[(x - ) / < (96.3 - 101.4) /82.4 ]
= 1 - P(z < -0.06)
Using z table,
= 1 -0.4761
=0.5239
(B)
n = 129
= 101.4
= / n = 82.4 / 129 = 7.2549
P( > 96.3) = 1 - P( < 96.3)
= 1 - P[( - ) / < (96.3 - 101.4) / 7.2549]
= 1 - P(z <-0.70 )
Using z table,
= 1 - 0.242
= 0.7580
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