Someone help please ://////
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Answer:
Mean :
a) Find the probability that an individual distance is greater than 204.30 cm.
We are supposed to find P(Z>204.30)
x = 204.30
Formula :
Substitute the value sin the formula :
P(Z>204.30)=1-P(Z<204.30)
Refer the z table
P(Z>1.13253)=1-P(Z<1.13253)
P(Z>1.13253)=1-0.8708
P(Z>1.13253)=0.1292
The probability that an individual distance is greater than 204.30 cm is 0.1292.
b) Find the probability that the mean for 20 randomly selected distances is greater than 192.80 cm
We are supposed to find P(Z>192.80)
x = 192.80
Formula :
Substitute the value sin the formula :
P(Z>192.80)=1-P(Z<192.80)
Refer the z table
P(Z>-0.265)=1-P(Z<-0.265)
P(Z>-0.265)=1-0.3974
P(Z>-0.265)=0.6026
The probability that the mean for 20 randomly selected distances is greater than 192.80 cm is 0.6026
c) Why can the normal distribution be used in part (b), even though the sample size does not exceed 30
The normal distribution is used because the original population has normal distribution.