Answer :
To determine the percentage of values within three standard deviations of the mean in a normal distribution, we can follow these steps:
1. Define the mean ([tex]$\mu$[/tex]) and standard deviation ([tex]$\sigma$[/tex]). For a standard normal distribution:
[tex]$$
\mu = 0 \quad \text{and} \quad \sigma = 1.
$$[/tex]
2. Calculate the lower and upper bounds corresponding to three standard deviations from the mean:
[tex]$$
\text{Lower bound} = \mu - 3\sigma = 0 - 3 \cdot 1 = -3,
$$[/tex]
[tex]$$
\text{Upper bound} = \mu + 3\sigma = 0 + 3 \cdot 1 = 3.
$$[/tex]
3. The probability that a value lies between these two bounds is given by the difference of the cumulative distribution function (CDF) evaluated at these points:
[tex]$$
P(-3 \leq X \leq 3) = \Phi(3) - \Phi(-3),
$$[/tex]
where [tex]$\Phi(x)$[/tex] is the standard normal CDF.
4. Evaluating these values, we have:
[tex]$$
\Phi(3) \approx 0.9986501 \quad \text{and} \quad \Phi(-3) \approx 0.0013499.
$$[/tex]
Therefore,
[tex]$$
P(-3 \leq X \leq 3) \approx 0.9986501 - 0.0013499 = 0.9973002.
$$[/tex]
5. To convert this probability to a percentage, multiply by 100:
[tex]$$
0.9973002 \times 100 \approx 99.73\%.
$$[/tex]
Thus, approximately [tex]$99.73\%$[/tex] of values in a normal distribution lie within three standard deviations of the mean.
The correct answer is [tex]$\boxed{99.7}$[/tex] (percent).
1. Define the mean ([tex]$\mu$[/tex]) and standard deviation ([tex]$\sigma$[/tex]). For a standard normal distribution:
[tex]$$
\mu = 0 \quad \text{and} \quad \sigma = 1.
$$[/tex]
2. Calculate the lower and upper bounds corresponding to three standard deviations from the mean:
[tex]$$
\text{Lower bound} = \mu - 3\sigma = 0 - 3 \cdot 1 = -3,
$$[/tex]
[tex]$$
\text{Upper bound} = \mu + 3\sigma = 0 + 3 \cdot 1 = 3.
$$[/tex]
3. The probability that a value lies between these two bounds is given by the difference of the cumulative distribution function (CDF) evaluated at these points:
[tex]$$
P(-3 \leq X \leq 3) = \Phi(3) - \Phi(-3),
$$[/tex]
where [tex]$\Phi(x)$[/tex] is the standard normal CDF.
4. Evaluating these values, we have:
[tex]$$
\Phi(3) \approx 0.9986501 \quad \text{and} \quad \Phi(-3) \approx 0.0013499.
$$[/tex]
Therefore,
[tex]$$
P(-3 \leq X \leq 3) \approx 0.9986501 - 0.0013499 = 0.9973002.
$$[/tex]
5. To convert this probability to a percentage, multiply by 100:
[tex]$$
0.9973002 \times 100 \approx 99.73\%.
$$[/tex]
Thus, approximately [tex]$99.73\%$[/tex] of values in a normal distribution lie within three standard deviations of the mean.
The correct answer is [tex]$\boxed{99.7}$[/tex] (percent).
