The average length of an adult blue whate is 30 meters with a standard deviation of 5.7 meters. If a sample of 31 whales is taken, what is the probability that their average lenght differs from the mean by more than 2.0 meters
Use z-scores using sample size:
z = (x - mean)/(sd/√n)
Find two scores:
z = (32 - 30)/(5.7/√31)
z = (28 - 30)/(5.7/√31)
Find the probability between the two z-scores using a z-table. Subtract that value from 1 for your answer.
I hope this will help get you started.
To find the probability that the sample's average length differs from the mean by more than 2.0 meters, we need to use the Central Limit Theorem and Z-scores.
The Central Limit Theorem states that if we have a large enough sample size, the sampling distribution of the sample mean will approach a normal distribution, regardless of the shape of the population distribution. In this case, we are given a sample of 31 whales, which is reasonably large.
First, we need to calculate the standard error of the mean (SE) using the formula:
SE = standard deviation / square root of sample size
SE = 5.7 / √31
SE ≈ 1.02
Next, we can calculate the Z-score using the formula:
Z = (sample mean - population mean) / SE
In this case, the value of the sample mean is 30 (since it is the same as the population mean). So the Z-score becomes:
Z = (30 - 30) / 1.02 = 0
Since we want to find the probability that the sample mean differs from the population mean by more than 2.0 meters, we need to find the probability of the absolute value of the Z-score being greater than 2.0.
Using a Z-table or a calculator, we can determine that the area under the standard normal curve for a Z-score greater than 2.0 or less than -2.0 is approximately 0.0455.
However, we are interested in the probability of it being either greater than 2.0 or less than -2.0 (i.e., in both tails). Thus, we multiply the result by 2:
Probability = 2 * 0.0455 = 0.091
Therefore, the probability that the average length of the sample of 31 whales differs from the mean by more than 2.0 meters is approximately 0.091, or 9.1%.