Answer :
a. The z-score for a temperature of 66 degrees in Lake Larson is 1.6.
b. The probability that the temperature of the lake will be greater than 66 degrees is approximately 0.0548, or 5.48%.
c. The temperature that separates the bottom 10% is approximately 52.4 degrees.
a. To find the z-score for a temperature of 66 degrees, we use the formula z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation. Plugging in the values, we get z = (66 - 58) / 5 = 1.6.
b. To find the probability that the temperature of the lake will be greater than 66 degrees, we can use the standard normal distribution table or a calculator. The z-score of 1.6 corresponds to a probability of approximately 0.0548, or 5.48%. This means that there is a 5.48% chance that the temperature of the lake will be greater than 66 degrees.
Interpreting the result, we can say that it is relatively uncommon for the temperature of Lake Larson to exceed 66 degrees. The majority of the temperature readings are expected to fall below 66 degrees, with only a small portion (approximately 5.48%) exceeding that value.
c. To find the temperature that separates the bottom 10%, we need to find the z-score corresponding to the 10th percentile of the standard normal distribution. This can be done using the z-table or a calculator. The z-score that corresponds to the 10th percentile is approximately -1.28. We can then use the z-score formula to find the temperature: -1.28 = (x - 58) / 5. Solving for x, we get x ≈ 52.4 degrees. Therefore, the temperature that separates the bottom 10% is approximately 52.4 degrees.
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