Suppose the masses for a particular population of adult Eastern chipmunks are normally distributed with a population mean of 110 grams and a population standard deviation of 12 grams. Explain whether 40% of the adult Eastern chipmunks have a mass that is less than or equal to 105.2 grams. Round to four decimal places as needed.
A) No. The z-score for a mass of 105.2 grams is −0.4, but 34.46%
of chipmunks have a mass that is less than or equal to 105.2
grams.
B) No. The z-score for a mass of 105.2 grams is −0.48, so 31.56%
of chipmunks have a mass that is less than or equal to 105.2
grams.
C) No. The z-score for a mass of 105.2 grams is −0.48, so 68.44%
of chipmunks have a mass that is less than or equal to 105.2
grams.
D) Yes. The z-score for a mass of 105.2 grams is −0.4, which means 40%
of chipmunks have a mass that is less than or equal to 105.2
grams.
To determine whether 40% of the adult Eastern chipmunks have a mass that is less than or equal to 105.2 grams, we need to calculate the z-score for a mass of 105.2 grams and find the corresponding percentage of chipmunks.
The z-score for a given value can be calculated using the formula:
z = (x - μ) / σ
Where x is the given value, μ is the population mean, and σ is the population standard deviation.
In this case, the given value is 105.2 grams, the population mean is 110 grams, and the population standard deviation is 12 grams.
Substituting these values into the formula, we get:
z = (105.2 - 110) / 12
z = -0.4
Now, we need to find the corresponding percentage of chipmunks for a z-score of -0.4.
According to the empirical rule or the normal distribution table, a z-score of -0.4 corresponds to approximately 34.46% of the population.
Therefore, the correct answer is A) No. The z-score for a mass of 105.2 grams is −0.4, but 34.46% of chipmunks have a mass that is less than or equal to 105.2 grams.