Final answer:
To determine the probability of a sample mean being less than 59.5 kg for chupacabras, we calculate the standard error, find the z-score for 59.5 kg, and then look up the probability for that z-score. The sample size is large enough to use the Central Limit Theorem, which allows us to treat the sampling distribution of the sample mean as normally distributed.
Step-by-step explanation:
To find the probability that a sample of 68 chupacabra from Cititon Forest have masses with a mean less than 59.5 kg, we will use the Central Limit Theorem. This theorem tells us that the sampling distribution of the sample mean is normally distributed if the sample size is large enough, even if the underlying population is not normally distributed. Given that we have a large sample size and the population standard deviation is known, this applies to our situation.
The mean of the sample means is equal to the population mean, which is 59.7 kg, and the standard deviation of the sample means (also known as the standard error) is the population standard deviation divided by the square root of the sample size.
First, we calculate the standard error:
Standard Error = Population Standard Deviation / sqrt(Sample Size) = 10.1 kg / sqrt(68) = 1.2247 kg
Next, we calculate the z-score for a sample mean of 59.5:
Z = (Sample Mean - Population Mean) / Standard Error = (59.5 kg - 59.7 kg) / 1.2247 kg = -0.1636
Finally, using a standard normal distribution table or a calculator, we find the probability corresponding to a z-score of -0.1636. This gives us the likelihood that the sample mean would be less than 59.5 kg.