Final answer:
Using the standard error of the sample mean for a normally distributed population of pyridoxine content and the z-score for the 5% left tail, 95% of all sample means will be greater than approximately 101.775 grams.
Step-by-step explanation:
To determine the value for which 95% of all sample means of pyridoxine content will be greater, we use the concept of the sampling distribution of the sample mean. Given that the population mean (μ) is 110 grams and the population standard deviation (σ) is 25 grams, and we are taking a sample size (n) of 25, we can compute the standard error of the mean.
The standard error of the mean (SEM) is given by σ / √n, which is 25 / √25, equal to 5 grams. According to the Central Limit Theorem, the distribution of sample means will be normally distributed with a mean equal to the population mean and a standard deviation equal to the SEM.
To find the cutoff value for the lower 5% of the distribution (because we want 95% of sample means to be greater than this value), we look up the z-score that corresponds to the 5% left tail of the standard normal distribution, which is approximately -1.645. We then use the formula:
x = μ + (z)(SEM)
x = 110 + (-1.645)(5) = 110 - 8.225 = 101.775 grams
Therefore, 95% of all the sample means will be greater than 101.775 grams.