Assuming that the true population mean is 98.6 degrees F and population standard deviation is 0.62 degrees F, what is the probability of obtaining a sample of 106 temperatures with a sample mean of 98.2 degrees F or less? Round the answer to four decimal places.
To find the probability of obtaining a sample mean of 98.2 degrees F or less, we need to use the standard deviation of the sampling distribution. The formula for finding the standard deviation (standard error) of the sample mean is:
σ/√n
Where:
- σ (population standard deviation) is 0.62 degrees F,
- n (sample size) is 106.
By substituting the values into the formula, we get:
0.62 / √106 ≈ 0.0604
We now need to find the Z-score for a sample mean of 98.2 degrees F. The Z-score formula for a sample mean is:
Z = (X - μ) / (σ/√n)
Where:
- X is the sample mean,
- μ is the true population mean, which is 98.6 degrees F,
- σ is the population standard deviation, which is 0.62 degrees F,
- n is the sample size, which is 106.
By substituting the values into the formula, we get:
Z = (98.2 - 98.6) / (0.62 / √106)
≈ (-0.4) / 0.0604
≈ -6.6225
Now, we can use the Z-table or a calculator to find the probability of obtaining a Z-score of -6.6225 or less. The Z-table provides the area under the standard normal curve.
Using the Z-table, we find that the area to the left of -6.6225 is virtually 0.
Therefore, the probability of obtaining a sample of 106 temperatures with a sample mean of 98.2 degrees F or less is approximately 0.
Keep in Mind: Since the Z-score is quite extreme in this case, the probability is extremely small.