Question

Suppose a variable has a normal distribution with mean 12 and standard deviation 2. Use the Empirical Rule to calculate the approximate PERCENTAGE area.

What is the PERCENTAGE of values ABOVE 10?

Note: Enter X.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 7 is entered as 7.00, 3.5 is entered as 3.50, 0.3750 is entered as 0.38

|Enter PERCENTAGE in above blank with NO % sign.

|

Suppose a variable has a normal distribution with mean 12 and standard deviation 2. Use the Empirical Rule to calculate the approximate PERCENTAGE area.

What is the PERCENTAGE of values BELOW 12?

Note: Enter X.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 27 is entered as 27.00, 3.5 is entered as 3.50, 0.3750 is entered as 0.38

|Enter PERCENTAGE in above blank with NO % sign.

|

Suppose a variable has a normal distribution with mean 12 and standard deviation 2. Use the Empirical Rule to calculate the approximate PERCENTAGE area.

What is the PERCENTAGE of values is contained within the interval (8, 14)?

Note: Enter X.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 27 is entered as 27.00, 3.5 is entered as 3.50, 0.3750 is entered as 0.38

|Enter PERCENTAGE in above blank with NO % sign.

|

Suppose a variable has a normal distribution with mean 12 and standard deviation 2. Use the Empirical Rule to calculate the approximate PERCENTAGE area.

What is the PERCENTAGE of values is contained BETWEEN 12 and 16?

Note: Enter X.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 27 is entered as 27.00, 3.5 is entered as 3.50, 0.3750 is entered as 0.38

|Enter PERCENTAGE in above blank with NO % sign.

|

Suppose a variable has a normal distribution with mean 12 and standard deviation 2. Use the Empirical Rule to calculate the approximate PERCENTAGE area.

What is the PERCENTAGE of values is contained within the interval (10, 14)?

Note: Enter X.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 27 is entered as 27.00, 3.5 is entered as 3.50, 0.3750 is entered as 0.38

Answer 1

From the graph, P(X>10) = 34%+34%+ 13.5%+2.35%+0.15% = 0.84

P(X<12) = 34%+13.5%+2.35%+0.15% = 0.50

P(8<X<14) = 13.5%+34%+ 34% = 0.815 = 0.82

P(12<X<16) = 34% + 13.5% = 0.48

P(10<X<14) = 34% + 34% = 0.68