Final answer:
The probability that the sample mean for the per capita consumption of milk will differ from the true mean by greater than 5.11 liters is approximately 0.0734 or 7.34%.
Step-by-step explanation:
To find the probability that the sample mean would differ from the true mean by greater than 5.11 liters, given the mean per capita consumption of milk per year is 153 liters with a standard deviation of 27 liters and a sample size of 90 people, we use the concept of the sampling distribution of the sample mean.
First, we calculate the standard error (SE) of the mean using the formula:
SE = \(\frac{\sigma}{\sqrt{n}}\) = \(\frac{27}{\sqrt{90}}\) = 2.85 liters
Next, we find the z-scores corresponding to the mean difference of 5.11 liters:
Z = \(\frac{X - \mu}{SE}\) = \(\frac{5.11}{2.85}\)
A Z of approximately 1.79 for the positive difference and -1.79 for the negative difference.
Using the Z-table, we find the probabilities corresponding to these z-scores and sum them up to get the total probability of the sample mean differing from the true mean by more than 5.11 liters:
P(Z > 1.79) + P(Z < -1.79)
To get a cumulative probability from a Z-table, we use the symmetry of the normal distribution. The probability for Z > 1.79 is 1 - P(Z < 1.79), and since the distribution is symmetric, P(Z < -1.79) = P(Z > 1.79).
From the Z-table, P(Z < 1.79) is approximately 0.9633, thus P(Z > 1.79) is 1 - 0.9633 = 0.0367. The total probability is:
P(differing by more than 5.11 liters) = 2 * 0.0367 = 0.0734
Therefore, the probability that the sample mean will differ from the true mean by greater than 5.11 liters is approximately 0.0734 or 7.34% when rounded to four decimal places.