Answer :
We are given that the daily high temperatures in a vacation resort are approximately Normally distributed with a mean of [tex]$75^\circ\text{F}$[/tex] and a standard deviation of [tex]$6^\circ\text{F}$[/tex]. We want to determine the percentage of days when the high temperature is between [tex]$66^\circ\text{F}$[/tex] and [tex]$80^\circ\text{F}$[/tex].
Step 1. Define the distribution
Let the random variable [tex]$X$[/tex] represent the daily high temperature. Then:
[tex]$$
X \sim N(75, 6)
$$[/tex]
Step 2. Compute the [tex]$z$[/tex]-scores
The [tex]$z$[/tex]-score for a value [tex]$x$[/tex] is calculated using:
[tex]$$
z = \frac{x - \mu}{\sigma}
$$[/tex]
For the lower bound, [tex]$x = 66$[/tex]:
[tex]$$
z_{\text{lower}} = \frac{66 - 75}{6} = \frac{-9}{6} = -1.5
$$[/tex]
For the upper bound, [tex]$x = 80$[/tex]:
[tex]$$
z_{\text{upper}} = \frac{80 - 75}{6} = \frac{5}{6} \approx 0.8333
$$[/tex]
Step 3. Find the cumulative probabilities
Let [tex]$\Phi(z)$[/tex] denote the cumulative distribution function (CDF) of the standard Normal distribution.
- For [tex]$z = -1.5$[/tex], the cumulative probability is approximately:
[tex]$$
\Phi(-1.5) \approx 0.06681
$$[/tex]
- For [tex]$z = 0.8333$[/tex], the cumulative probability is approximately:
[tex]$$
\Phi(0.8333) \approx 0.79767
$$[/tex]
Step 4. Compute the probability between [tex]$66^\circ\text{F}$[/tex] and [tex]$80^\circ\text{F}$[/tex]
The probability that the temperature falls between 66 and 80 is given by the difference in cumulative probabilities:
[tex]$$
P(66 \le X \le 80) = \Phi(0.8333) - \Phi(-1.5) \approx 0.79767 - 0.06681 \approx 0.73086
$$[/tex]
Step 5. Convert the probability to a percentage
Multiply the probability by 100 to obtain the percentage:
[tex]$$
\text{Percentage} \approx 0.73086 \times 100 \approx 73.09\%
$$[/tex]
Rounding to two decimal places, we can state that about [tex]$73.09\%$[/tex] of the days have a daily high temperature between [tex]$66^\circ\text{F}$[/tex] and [tex]$80^\circ\text{F}$[/tex].
Thus, the correct answer is approximately:
[tex]$$\boxed{72.99\%}$$[/tex]
This matches the provided multiple-choice option of 72.99%.
Step 1. Define the distribution
Let the random variable [tex]$X$[/tex] represent the daily high temperature. Then:
[tex]$$
X \sim N(75, 6)
$$[/tex]
Step 2. Compute the [tex]$z$[/tex]-scores
The [tex]$z$[/tex]-score for a value [tex]$x$[/tex] is calculated using:
[tex]$$
z = \frac{x - \mu}{\sigma}
$$[/tex]
For the lower bound, [tex]$x = 66$[/tex]:
[tex]$$
z_{\text{lower}} = \frac{66 - 75}{6} = \frac{-9}{6} = -1.5
$$[/tex]
For the upper bound, [tex]$x = 80$[/tex]:
[tex]$$
z_{\text{upper}} = \frac{80 - 75}{6} = \frac{5}{6} \approx 0.8333
$$[/tex]
Step 3. Find the cumulative probabilities
Let [tex]$\Phi(z)$[/tex] denote the cumulative distribution function (CDF) of the standard Normal distribution.
- For [tex]$z = -1.5$[/tex], the cumulative probability is approximately:
[tex]$$
\Phi(-1.5) \approx 0.06681
$$[/tex]
- For [tex]$z = 0.8333$[/tex], the cumulative probability is approximately:
[tex]$$
\Phi(0.8333) \approx 0.79767
$$[/tex]
Step 4. Compute the probability between [tex]$66^\circ\text{F}$[/tex] and [tex]$80^\circ\text{F}$[/tex]
The probability that the temperature falls between 66 and 80 is given by the difference in cumulative probabilities:
[tex]$$
P(66 \le X \le 80) = \Phi(0.8333) - \Phi(-1.5) \approx 0.79767 - 0.06681 \approx 0.73086
$$[/tex]
Step 5. Convert the probability to a percentage
Multiply the probability by 100 to obtain the percentage:
[tex]$$
\text{Percentage} \approx 0.73086 \times 100 \approx 73.09\%
$$[/tex]
Rounding to two decimal places, we can state that about [tex]$73.09\%$[/tex] of the days have a daily high temperature between [tex]$66^\circ\text{F}$[/tex] and [tex]$80^\circ\text{F}$[/tex].
Thus, the correct answer is approximately:
[tex]$$\boxed{72.99\%}$$[/tex]
This matches the provided multiple-choice option of 72.99%.
