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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.
Answer: C) No, because one x-value corresponds to two different y-values.
Explanation:
At the bottom of the table, we see that x = 6 corresponds to y = 5 and y = 6 at the same time. Such a thing is not possible if we want a function to happen. A function is only valid when any x input leads to exactly one y output (assuming this x value is in the domain). We'd have to cross off either of these two rows to help make this table a function.
As a visual approach, you can plot each row as (x,y) points on the same coordinate grid. Then notice how (6,5) and (6,6) line up vertically, which means this collection of points fail the vertical line test, and this is a visual way to see why we don't have a function.
