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Sampling Distribution of a Sample Mean: Get to Class!

APSTAT-Y3XVYF

During passing time at a high school, it takes a student an average of $4.1\text{ minutes}$ with a standard deviation of $0.88\text{ minutes}$ to make it from one class to another class.

If the students have $5\text{ minutes}$ to make it from one class to another class without being late, what is the probability when $100\text{ students}$ are randomly selected their average time will be GREATER than $5\text{ minutes}$?

A

$P \left( z >\cfrac { 4.1-5 }{ \sqrt { \frac { (0.88)(0.12) }{ 100 } } } \right)$

B

$P \left( z >\cfrac { 4.1-5 }{ \frac { 0.88 }{ \sqrt { 100 } } } \right)$

C

$P \left( z >\cfrac { 5-4.1 }{ 0.88 } \right)$

D

$P \left( z >\cfrac { 5-4.1 }{ \sqrt { \frac { (0.88)(0.12) }{ 100 } } } \right)$

E

$P \left( z >\cfrac { 5-4.1 }{ \frac { 0.88 }{ \sqrt { 100 } } } \right)$