Answer :
To solve the problems given, we need to understand the properties of a normal distribution and use the z-score formula. The mean (\mu\u00a0) of the distribution is 97.6, and the standard deviation (\sigma\u00a0) is 3.5.
Finding the 78th Percentile:
Percentiles indicate the value below which a given percentage of observations fall. To find the 78th percentile, we need to determine the z-score that corresponds to 0.78 in the standard normal distribution table.
For a z-score of approximately 0.772, the formula used is:
[
x = \mu + z \times \sigma
]
[tex]x = 97.6 + 0.772 \times 3.5 = 100.292[/tex]
Thus, the temperature that is the 78th percentile of this distribution is approximately 100.3. The answer is A) 100.3.
Probability of a Specific Temperature:
To find the probability of a temperature of 99, we use:
[
z = \frac{99 - 97.6}{3.5} \approx 0.4
]
Checking z-tables or using a calculator, the probability for z = 0.4 is about 0.6554.
The probability that a summer day has a high of exactly 99 is approximately 0 because we deal with continuous distributions. Specific values have a probability of 0.
Probability of Temperature Greater Than 99:
First, find the cumulative probability:
Since the z-score for 99 was found to be 0.4, and the cumulative probability of z = 0.4 is 0.6554, the probability of the temperature being greater than 99 is:
[
p( x > 99 ) = 1 - 0.6554 = 0.3446
]
Thus, the probability that a randomly selected summer day has a temperature greater than 99 is C) 0.3446.
Average Temperature for 15 Days Greater Than 99:
When dealing with the average of several observations, we use the central limit theorem. The means of samples will have a normal distribution with:
[
\mu_{\text{sample}} = \mu = 97.6
]
[tex]\sigma_{\text{sample}} = \frac{\sigma}{\sqrt{n}} = \frac{3.5}{\sqrt{15}} \approx 0.903[/tex]
Calculate the z-score for an average temperature of 99:
[tex]z = \frac{99 - 97.6}{0.903} \approx 1.548[/tex]
Find this cumulative probability for z = 1.548 which is about 0.9394, and therefore:
[tex]p( \bar{x} > 99 ) = 1 - 0.9394 = 0.0606[/tex]
This means the probability that the average temperature of 15 days is greater than 99 is A) 0.0606.