Answer :
The probability that a single randomly selected value from the population is greater than 59.1 is approximately 0.3446.
To find this probability, we use the standard normal distribution and the z-score formula. First, we calculate the z-score for the value 59.1 using the formula z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation. Substituting the given values, we find z ≈ (59.1 - 41.1) / 89.5 ≈ 0.2011. Next, we use a standard normal distribution table or calculator to find the probability corresponding to this z-score. The probability that a z-score is greater than 0.2011 is approximately 0.3446.
Step by step explanation:
1. Calculate the z-score: z = (59.1 - 41.1) / 89.5 ≈ 0.2011.
2. Use a standard normal distribution table or calculator to find the probability corresponding to a z-score of 0.2011.
3. The probability that a z-score is greater than 0.2011 is approximately 0.3446.
4. Therefore, the probability that a single randomly selected value from the population is greater than 59.1 is approximately 0.3446.
Thus, the final answer is that the probability of selecting a value greater than 59.1 from the population is approximately 0.3446, rounded to four decimal places.
