Suppose we want to study the average weight of children in the US. We collect a random sample of 345 first graders in three diff
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Answer:
2.87%
Step-by-step explanation:
We have the following information:
mean (m) = 200
standard deviation (sd) = 50
sample size = n = 40
the probability that their mean is above 21.5 is determined as follows:
P (x> 21.5) = P [(x - m) / (sd / n ^ (1/2))> (21.5 - 200) / (50/40 ^ (1/2))]
P (x> 21.5) = P (z> -22.57)
this value is very strange, therefore I suggest that it is not 21.5 but 215, therefore it would be:
P (x> 215) = P [(x - m) / (sd / n ^ (1/2))> (215 - 200) / (50/40 ^ (1/2))]
P (x> 215) = P (z> 1.897)
P (x> 215) = 1 - P (z <1.897)
We look for this value in the attached table of z and we have to:
P (x> 215) = 1 - 0.9713 (attached table)
P (x> 215) =.0287
Therefore the probability is approximately 2.87%
