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Answer :
- Calculate the z-scores for $55^{\circ} F$ and $76^{\circ} F$: $z_1 = -2$ and $z_2 = 1$.
- Find the probabilities corresponding to the z-scores: $P(Z \le 1) \approx 0.8413$ and $P(Z \le -2) \approx 0.0228$.
- Calculate the probability $P(-2 \le Z \le 1) = P(Z \le 1) - P(Z \le -2) = 0.8413 - 0.0228 = 0.8185$.
- Convert to percentage: $0.8185 \times 100 = \boxed{81.5 \%}$.
### Explanation
1. Understand the problem and provided data
We are given that the average summer temperature in Anchorage is $69^{\circ} F$ and the standard deviation is $7^{\circ} F$. We want to find the percentage of time the temperature is between $55^{\circ} F$ and $76^{\circ} F$. Let $X$ be the daily temperature. Then $X \sim N(69, 7^2)$.
2. Calculate the z-scores
First, we need to convert the temperatures to z-scores using the formula $z = \frac{x - \mu}{\sigma}$. For $x_1 = 55$, we have $z_1 = \frac{55 - 69}{7} = \frac{-14}{7} = -2$. For $x_2 = 76$, we have $z_2 = \frac{76 - 69}{7} = \frac{7}{7} = 1$.
3. Define the probability
Now we want to find the probability that the temperature is between $55^{\circ} F$ and $76^{\circ} F$, which is equivalent to finding the probability that the z-score is between -2 and 1. That is, we want to find $P(-2 \le Z \le 1)$, where $Z$ is a standard normal random variable.
4. Find the probabilities corresponding to the z-scores
We can find this probability by looking up the values in a standard normal distribution table or using a calculator. We have $P(Z \le 1) \approx 0.8413$ and $P(Z \le -2) \approx 0.0228$.
5. Calculate the probability
Therefore, $P(-2 \le Z \le 1) = P(Z \le 1) - P(Z \le -2) = 0.8413 - 0.0228 = 0.8185$.
6. Convert to percentage and conclude
To convert this probability to a percentage, we multiply by 100: $0.8185 \times 100 = 81.85 \%$. The closest answer choice is $81.5 \%$.
### Examples
Understanding normal distributions and calculating probabilities is crucial in many real-world scenarios. For instance, in manufacturing, it helps determine the percentage of products that fall within acceptable quality ranges. In finance, it's used to model stock returns and assess risk. In weather forecasting, like this problem, it helps predict the likelihood of temperatures falling within a certain range, aiding in planning and resource allocation.
- Find the probabilities corresponding to the z-scores: $P(Z \le 1) \approx 0.8413$ and $P(Z \le -2) \approx 0.0228$.
- Calculate the probability $P(-2 \le Z \le 1) = P(Z \le 1) - P(Z \le -2) = 0.8413 - 0.0228 = 0.8185$.
- Convert to percentage: $0.8185 \times 100 = \boxed{81.5 \%}$.
### Explanation
1. Understand the problem and provided data
We are given that the average summer temperature in Anchorage is $69^{\circ} F$ and the standard deviation is $7^{\circ} F$. We want to find the percentage of time the temperature is between $55^{\circ} F$ and $76^{\circ} F$. Let $X$ be the daily temperature. Then $X \sim N(69, 7^2)$.
2. Calculate the z-scores
First, we need to convert the temperatures to z-scores using the formula $z = \frac{x - \mu}{\sigma}$. For $x_1 = 55$, we have $z_1 = \frac{55 - 69}{7} = \frac{-14}{7} = -2$. For $x_2 = 76$, we have $z_2 = \frac{76 - 69}{7} = \frac{7}{7} = 1$.
3. Define the probability
Now we want to find the probability that the temperature is between $55^{\circ} F$ and $76^{\circ} F$, which is equivalent to finding the probability that the z-score is between -2 and 1. That is, we want to find $P(-2 \le Z \le 1)$, where $Z$ is a standard normal random variable.
4. Find the probabilities corresponding to the z-scores
We can find this probability by looking up the values in a standard normal distribution table or using a calculator. We have $P(Z \le 1) \approx 0.8413$ and $P(Z \le -2) \approx 0.0228$.
5. Calculate the probability
Therefore, $P(-2 \le Z \le 1) = P(Z \le 1) - P(Z \le -2) = 0.8413 - 0.0228 = 0.8185$.
6. Convert to percentage and conclude
To convert this probability to a percentage, we multiply by 100: $0.8185 \times 100 = 81.85 \%$. The closest answer choice is $81.5 \%$.
### Examples
Understanding normal distributions and calculating probabilities is crucial in many real-world scenarios. For instance, in manufacturing, it helps determine the percentage of products that fall within acceptable quality ranges. In finance, it's used to model stock returns and assess risk. In weather forecasting, like this problem, it helps predict the likelihood of temperatures falling within a certain range, aiding in planning and resource allocation.
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