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Answer :
The probability that the sample mean will be less than 200 pounds is approximately 0.9015 or 90.15%.
Step-by-step explanation:
To solve this problem, we can use the central limit theorem, which states that the sampling distribution of the mean of a large sample (n >= 30) will be approximately normal, regardless of the distribution of the population.
We are given that the population mean is 196 pounds and the standard deviation is 22 pounds. The sample size is 50, which is large enough to apply the central limit theorem.
To find the probability that the mean of the sample will be less than 200 pounds, we need to standardize the sample mean using the formula:
z = (x - mu) / (sigma / sqrt(n))
where x is the sample mean, mu is the population mean, sigma is the population standard deviation, and n is the sample size.
Plugging in the values we get:
z = (200 - 196) / (22 / sqrt(50)) = 1.59
Now we need to find the probability that a standard normal variable is less than 1.59. We can use a standard normal table or calculator to find this probability, which is approximately 0.944.
Therefore, the probability that the mean of the sample will be less than 200 pounds is 0.944 or 94.4% (rounded to one decimal place).
So, we can say that there is a 94.4% chance that the mean weight of 50 randomly selected individuals will be less than 200 pounds.
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