The overhead reach distances of adult females are normally distributed with a mean of 202.5 cm and a standard deviation of 8 cm.
a. Find the probability that an individual distance is greater than 215.00 cm.
b. Find the probability that the mean for 15 randomly selected distances is greater than 200.30 cm.
c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?
a)
for normal distribution z score =(X-μ)/σx | |
mean μ= | 202.5 |
standard deviation σ= | 8 |
probability that an individual distance is greater than 215.00 cm :
probability =P(X>215)=P(Z>(215-202.5)/8)=P(Z>1.56)=1-P(Z<1.56)=1-0.9406=0.0591 |
b)
std error=σx̅=σ/√n=8/√15 = | 2.0656 |
probability that the mean for 15 randomly selected distances is greater than 200.30 cm :
probability =P(X>200.3)=P(Z>(200.3-202.5)/2.066)=P(Z>-1.07)=1-P(Z<-1.07)=1-0.1423=0.8566 |
c)
since population is normal ; therefore sampling distribution of sample mean is normal for any sample size and we do not require sample size to be larger than 30
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