This is a Question of Statistics and an application of standard normal distribution.
Average amount(Mean) of protein produced by the plant = 109.9
Standard Deviation = 8.1
Size of the sample = 60
Average amount of protein for the sample is the same as that of population.So average amount of protein for the sample = 109.9
Standard Deviation of the sample = (Population Standard Deviation) divided by (square root of sample size)
So sample standard deviation =
So, we are to find the probability of amount of protein being produced less than 107 for mean 109.9 and standard deviation 1.046.
First we convert this value to z score.
Z-score =
Using z-table, converting this z score to probability we get:
Probability of mean being less than 107 = 0.0028
Since the probability of mean being less than 107 is less than 0.05. This event is highly unlikely.
Average amount(Mean) of protein produced by the plant = 109.9
Standard Deviation = 8.1
Size of the sample = 60
Average amount of protein for the sample is the same as that of population.So average amount of protein for the sample = 109.9
Standard Deviation of the sample = (Population Standard Deviation) divided by (square root of sample size)
So sample standard deviation =
So, we are to find the probability of amount of protein being produced less than 107 for mean 109.9 and standard deviation 1.046.
First we convert this value to z score.
Z-score =
Using z-table, converting this z score to probability we get:
Probability of mean being less than 107 = 0.0028
Since the probability of mean being less than 107 is less than 0.05. This event is highly unlikely.