Question 16.
The 68—95—99.7% rule tells us approximately the percentage of a normal distribution which lies within one, two or three standard deviations of the mean. If 
Z
 is the standard normal random variable find, correct to two decimal places:
a.
a
 such that 
Pr
(
−
a
<
Z
<
a
)
=
0.68
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+ Check answer
b.
b
 such that 
Pr
(
−
b
<
Z
<
b
)
=
0.95
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+ Check answer
c.
c
 such that 
Pr
(
−
c
<
Z
<
c
)
=
0.997
+ Workspace
+ Check answer
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The 68–95–99.7% rule is tied to the empirical rule for normal distributions. It states that:

- **68%** of the distribution lies within **1 standard deviation** of the mean,
- **95%** lies within **2 standard deviations**,
- **99.7%** lies within **3 standard deviations**.

The random variable \( Z \) is from the standard normal distribution, which has a mean of 0 and a standard deviation of 1. Let’s solve each part of the question step by step.

### a. Finding \( a \) such that \( Pr(-a < Z < a) = 0.68 \)
The probability \( Pr(-a < Z < a) = 0.68 \) means that 68% of the values lie between \( -a \) and \( a \). From the 68-95-99.7 rule, we know that this corresponds to **1 standard deviation** on either side of the mean.

Thus, \( a = 1 \).

So, correct to two decimal places:
\[
a = 1.00
\]

### b. Finding \( b \) such that \( Pr(-b < Z < b) = 0.95 \)
For \( Pr(-b < Z < b) = 0.95 \), 95% of the distribution lies between \( -b \) and \( b \). From the empirical rule, this corresponds to **2 standard deviations** on either side of the mean.

Thus, \( b = 2 \).

So, correct to two decimal places:
\[
b = 2.00
\]

### c. Finding \( c \) such that \( Pr(-c < Z < c) = 0.997 \)
For \( Pr(-c < Z < c) = 0.997 \), 99.7% of the distribution lies between \( -c \) and \( c \). According to the empirical rule, this corresponds to **3 standard deviations**.

Thus, \( c = 3 \).

So, correct to two decimal places:
\[
c = 3.00
\]

### Final Answers:
- a. \( a = 1.00 \)
- b. \( b = 2.00 \)
- c. \( c = 3.00 \)

Would you like more details on any of these steps?

### Relative Questions:
1. What is the empirical rule and how is it derived?
2. How does the standard normal distribution differ from other normal distributions?
3. Why do we use the standard normal distribution in statistics?
4. Can you explain the role of Z-scores in normal distribution problems?
5. How can the empirical rule help in determining probabilities for other distributions?

**Tip**: The empirical rule is most useful when the data approximates a normal distribution, making it a quick tool for estimating probabilities without needing a detailed Z-table.

The 68—95—99.7% rule tells us approximately the percentage of a normal distribution which lies within one, two or three standard deviations of the mean. If Z is the standard normal random variable find, correct to two decimal places:
a. a such that Pr(−a < Z < a) = 0.68
b. b such that Pr(−b < Z < b) = 0.95
c. c such that Pr(−c < Z < c) = 0.997

Understand how to solve for standard deviations in a normal distribution using the 68-95-99.7 empirical rule. This question involves finding values of a, b, and c such that Pr(−a < Z < a) = 0.68, Pr(−b < Z < b) = 0.95, and Pr(−c < Z < c) = 0.997. A step-by-step solution is provided, perfect for high school and early college students studying statistics.

Math Problem Statement

Question 16. The 68—95—99.7% rule tells us approximately the percentage of a normal distribution which lies within one, two or three standard deviations of the mean. If Z is the standard normal random variable find, correct to two decimal places: a. a such that Pr ( − a < Z < a )

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