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Answer :
To solve the problems given, we apply concepts from the normal distribution.
67th Percentile of the Distribution
To find the temperature that represents the 67th percentile, we need to find the corresponding z-score for the 67th percentile.
- The z-score for the 67th percentile is approximately 0.44 (you can find this using a standard normal distribution table or a calculator).
Using the z-score formula:
[tex]X = \mu + z \cdot \sigma[/tex]
where [tex]\mu = 99.8[/tex] (mean) and [tex]\sigma = 4.2[/tex] (standard deviation).
[tex]X = 99.8 + 0.44 \cdot 4.2 = 101.648 \approx 101.6[/tex]
Therefore, the summer high temperature that is the 67th percentile is approximately 101.6 (Option D).
Probability of a High Temperature of 102
Since temperatures are continuous, the probability of a temperature being exactly 102 is essentially 0 due to the nature of continuous distributions.
Therefore, the probability is 0 (Option B).
Probability of a Temperature Greater than 102
First, calculate the z-score for 102:
[tex]z = \frac{X - \mu}{\sigma} = \frac{102 - 99.8}{4.2} \approx 0.52[/tex]
Using the z-score table or a normal distribution calculator, the probability (P) for [tex]z \leq 0.52[/tex] is approximately 0.6985.
Therefore, the probability of [tex]z > 0.52[/tex] is:
[tex]1 - P(z \leq 0.52) = 1 - 0.6985 = 0.3015[/tex]
Hence, the probability is 0.3015 (Option D).
Probability of a Group of 25 Days with Average Greater than 102
For a sample mean, the standard deviation is different. It’s called the standard error, given by:
[tex]\text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{4.2}{\sqrt{25}} = 0.84[/tex]
Calculating the z-score for the average temperature:
[tex]z = \frac{102 - 99.8}{0.84} \approx 2.62[/tex]
Using the z-score table, the probability (P) for [tex]z \leq 2.62[/tex] is approximately 0.9956.
Therefore, the probability of [tex]z > 2.62[/tex] is:
[tex]1 - 0.9956 = 0.0044[/tex]
Thus, the probability is 0.0044 (Option D).
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