A population of values has a normal distribution with mean of
50.3 and standard deviation of 84.
Find the probability that from a sample of 226 the sample mean is
greater than 49.7.
Enter your answers as numbers accurate to 4 decimal places.
Solution :
Given that ,
mean = = 50.3
standard deviation = = 84
n = 226
= 50.3
= / n = 84/ 226 = 5.5876
P( >49.7 ) = 1 - P( <49.7 )
= 1 - P[( - ) / < (49.7 -50.3) / 5.5876]
= 1 - P(z <-0.11 )
Using z table
= 1 - 0.4562
= 0.5438
probability= 0.5438
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