The following TI-84 Plus display presents some population parameters. Assume the distribution is bell-shaped.

Question

x= 116
Ex = 14508
Ex^2= 1798992
Sx= 5.04310345
ox=5
n= 117
minX =118
down arrow Qv1 = 119

) Using the Empirical Rule, approximately what percentage of the population values are between 111 and 121?

Answer 1

The Empirical Rule, also known as the 68-95-99.7 rule, applies to bell-shaped (normal) distributions. It states that:

• Approximately 68% of the data fall within one standard deviation of the mean.
• Approximately 95% of the data fall within two standard deviations of the mean.
• Approximately 99.7% of the data fall within three standard deviations of the mean.

Given the information:

• Mean (\( \mu \)) \( x = 116 \)
• Standard deviation (\( \sigma \)) \( o_x = 5 \)

To find the percentage of population values between 111 and 121:

1. Calculate the number of standard deviations from the mean:

   • The lower limit (111) is: \[ z = \frac{111 - 116}{5} = \frac{-5}{5} = -1 \]
   • The upper limit (121) is: \[ z = \frac{121 - 116}{5} = \frac{5}{5} = 1 \]

2. Apply the Empirical Rule:

   • According to the Empirical Rule, approximately 68% of the data falls within one standard deviation of the mean (between \( \mu - \sigma \) and \( \mu + \sigma \)).

Thus, the approximate percentage of the population values between 111 and 121 is 68%.

Answer 2

Using the Empirical Rule, between what two values will approximately 95% of the population be?

Answer 3

Using the Empirical Rule, we can determine the interval that contains approximately 95% of the population values in a bell-shaped distribution.

Given:

• Mean (\( \mu \)) = 116
• Standard deviation (\( \sigma \)) = 5

To find the interval for approximately 95% of the population:

1. Determine the range of values that are within two standard deviations from the mean. According to the Empirical Rule, about 95% of the data falls within two standard deviations of the mean:

\[ \text{Lower limit} = \mu - 2\sigma = 116 - 2(5) = 116 - 10 = 106 \]

\[ \text{Upper limit} = \mu + 2\sigma = 116 + 2(5) = 116 + 10 = 126 \]

Conclusion:

Approximately 95% of the population values will be between 106 and 126.