Answer :
The probability that the sample mean of a sample of size 89 elements selected from this population is 0.5.
The probability that the sample mean of a sample of size 89 elements selected from a population with a mean of 98.2 and a standard deviation of 25.2 can be calculated using the Central Limit Theorem.
According to the Central Limit Theorem, when the sample size is sufficiently large (typically greater than or equal to 30) and the population is not extremely skewed, the distribution of sample means will be approximately normally distributed, regardless of the shape of the original population.
To find the probability, we need to convert the sample mean to a z-score and then use the z-table to determine the probability.
The formula for calculating the z-score is:
z = (x - μ) / (σ / sqrt(n))
where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.
In this case, x = 98.2, μ = 98.2, σ = 25.2, and n = 89.
Plugging these values into the formula, we get:
z = (98.2 - 98.2) / (25.2 / sqrt(89))
Simplifying the equation, we get:
z = 0 / 2.6637
Since the numerator is 0, the z-score is 0.
To find the probability, we can look up the z-score of 0 in the z-table.
The z-score of 0 corresponds to a probability of 0.5.
Therefore, the probability that the sample mean of a sample of size 89 elements selected from this population is 0.5.
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