On the sixth day of Statistics, my teacher gave to me, six geese-a-laying. The eggs they were a-laying were normally distributed and had a mean of 7cm and standard deviation of 2cm. What is the probability that if you randomly select one egg from each of the six geese, the mean of all six will be less than 6 cm?
In order to receive ANY credit, you must show formula, substitution
Z = (score-mean)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
what would I put for score ?
To find the probability that the mean of all six eggs will be less than 6 cm, we need to use the concept of the sampling distribution of the mean.
The sampling distribution of the mean is the distribution of all possible sample means if we were to take multiple samples from the same population. The central limit theorem states that when the sample size is large enough, the sampling distribution of the mean will be approximately normal, regardless of the shape of the original population distribution.
In this case, we have the mean and standard deviation of a single egg from the population, which are 7 cm and 2 cm respectively. Since we know the sample size (n) is 6, we can use the formula for the standard deviation of the sampling distribution of the mean:
Standard deviation of the sampling distribution of the mean = standard deviation of the population / square root of sample size
Plugging in the values:
Standard deviation of the sampling distribution of the mean = 2 cm / sqrt(6)
Now we need to calculate the z-score for the desired mean of 6 cm:
z-score = (desired mean - population mean) / standard deviation of the sampling distribution of the mean
= (6 - 7) / (2 / sqrt(6))
Calculating this, we get:
z-score = -1 / (2 / sqrt(6))
Finally, we need to find the probability of the z-score being less than this value. We can use a standard normal distribution table or a calculator to find this probability.