Final Answer:
The probability that the mean life expectancy will be less than 71.4 years for a sample of 50 people from this region is approximately 0.0951.
Step-by-step explanation:
The first step is to calculate the standard error of the mean (SEM), which is the standard deviation divided by the square root of the sample size. In this case, SEM = 5.3 / √50 ≈ 0.7496. Next, we calculate the z-score, which measures how many standard deviations a particular value is from the mean. The formula for the z-score is (X - μ) / SEM, where X is the value we're interested in, μ is the mean, and SEM is the standard error of the mean. For this problem, the z-score is (71.4 - 72.0) / 0.7496 ≈ -0.8001.
Now, we look up this z-score in the standard normal distribution table to find the probability. A z-score of -0.80 corresponds to a cumulative probability of approximately 0.2119. To find the probability that the mean life expectancy is less than 71.4 years, we subtract this value from 0.5 (since the standard normal distribution is symmetric). Therefore, P(Xˉ < 71.4) = 0.5 - 0.2119 ≈ 0.2881.
However, we need to consider that we rounded the z-score to two decimal places, which introduces some error. To correct for this, we adjust the probability using continuity correction. In this case, we add half the width of the interval to the left of 71.4. The final probability is approximately 0.2881 + 0.0050 ≈ 0.2931, rounded to four decimal places. Therefore, the probability that the mean life expectancy will be less than 71.4 years is approximately 0.0951.