Answer :
To solve the problem of finding the probability that an individual had between \[tex]$0.71 and \$[/tex]0.92, given a normal distribution with [tex]\(\mu = 1\)[/tex] and [tex]\(\sigma = 1\)[/tex], we'll proceed as follows:
### Step-by-Step Solution
1. Determine the z-scores:
The z-score for a value [tex]\(x\)[/tex] in a normal distribution is calculated using the formula:
[tex]\[ z = \frac{x - \mu}{\sigma} \][/tex]
For the lower bound [tex]\( x = 0.71 \)[/tex]:
[tex]\[ z_{\text{lower}} = \frac{0.71 - 1}{1} = -0.29 \][/tex]
For the upper bound [tex]\( x = 0.92 \)[/tex]:
[tex]\[ z_{\text{upper}} = \frac{0.92 - 1}{1} = -0.08 \][/tex]
2. Find the cumulative probabilities:
Using the standard normal distribution table (or a cumulative distribution function, CDF):
- The cumulative probability [tex]\( P(Z < z_{\text{lower}}) \)[/tex] for [tex]\( z_{\text{lower}} = -0.29 \)[/tex] is approximately 0.3859.
- The cumulative probability [tex]\( P(Z < z_{\text{upper}}) \)[/tex] for [tex]\( z_{\text{upper}} = -0.08 \)[/tex] is approximately 0.4681.
3. Calculate the probability between the two z-scores:
To find the probability that an individual has between [tex]$0.71 and $[/tex]0.92, we subtract the cumulative probabilities:
[tex]\[ P(0.71 < X < 0.92) = P(Z < z_{\text{upper}}) - P(Z < z_{\text{lower}}) \][/tex]
[tex]\[ P(0.71 < X < 0.92) = 0.4681 - 0.3859 = 0.0822 \][/tex]
Thus, the probability that an individual had between \[tex]$0.71 and \$[/tex]0.92 is 0.0822 (rounded to four decimal places).
### Graphing the Situation
To visualize this, you would draw the standard normal distribution curve, which is bell-shaped and centered at [tex]\(z = 0\)[/tex].
- Shade the area between [tex]\(z = -0.29\)[/tex] and [tex]\(z = -0.08\)[/tex].
Let's break this down:
1. Draw a horizontal axis (z-axis) and a vertical axis (probability density).
2. Sketch the bell-shaped curve of the standard normal distribution.
3. Mark [tex]\(z = -0.29\)[/tex] and [tex]\(z = -0.08\)[/tex] on the z-axis.
4. Shade the area under the curve between these two z-scores.
This shaded region represents the probability that an individual had between \[tex]$0.71 and \$[/tex]0.92, which is 0.0822.
### Step-by-Step Solution
1. Determine the z-scores:
The z-score for a value [tex]\(x\)[/tex] in a normal distribution is calculated using the formula:
[tex]\[ z = \frac{x - \mu}{\sigma} \][/tex]
For the lower bound [tex]\( x = 0.71 \)[/tex]:
[tex]\[ z_{\text{lower}} = \frac{0.71 - 1}{1} = -0.29 \][/tex]
For the upper bound [tex]\( x = 0.92 \)[/tex]:
[tex]\[ z_{\text{upper}} = \frac{0.92 - 1}{1} = -0.08 \][/tex]
2. Find the cumulative probabilities:
Using the standard normal distribution table (or a cumulative distribution function, CDF):
- The cumulative probability [tex]\( P(Z < z_{\text{lower}}) \)[/tex] for [tex]\( z_{\text{lower}} = -0.29 \)[/tex] is approximately 0.3859.
- The cumulative probability [tex]\( P(Z < z_{\text{upper}}) \)[/tex] for [tex]\( z_{\text{upper}} = -0.08 \)[/tex] is approximately 0.4681.
3. Calculate the probability between the two z-scores:
To find the probability that an individual has between [tex]$0.71 and $[/tex]0.92, we subtract the cumulative probabilities:
[tex]\[ P(0.71 < X < 0.92) = P(Z < z_{\text{upper}}) - P(Z < z_{\text{lower}}) \][/tex]
[tex]\[ P(0.71 < X < 0.92) = 0.4681 - 0.3859 = 0.0822 \][/tex]
Thus, the probability that an individual had between \[tex]$0.71 and \$[/tex]0.92 is 0.0822 (rounded to four decimal places).
### Graphing the Situation
To visualize this, you would draw the standard normal distribution curve, which is bell-shaped and centered at [tex]\(z = 0\)[/tex].
- Shade the area between [tex]\(z = -0.29\)[/tex] and [tex]\(z = -0.08\)[/tex].
Let's break this down:
1. Draw a horizontal axis (z-axis) and a vertical axis (probability density).
2. Sketch the bell-shaped curve of the standard normal distribution.
3. Mark [tex]\(z = -0.29\)[/tex] and [tex]\(z = -0.08\)[/tex] on the z-axis.
4. Shade the area under the curve between these two z-scores.
This shaded region represents the probability that an individual had between \[tex]$0.71 and \$[/tex]0.92, which is 0.0822.
