Question

thanks to social media, the happiest creature on earth, an always smiling australian marsupial called a quokka, has become well - known. suppose that weights of quokkas can be modeled with a normal model with a mean of 7 pounds and a standard deviation of 1.8 pounds.
a) how many standard deviations from the mean would a quokka weighing 3 pounds be?
b) which would be more unusual, a quokka weighing 3 pounds or one weighing 9 pounds?
a) a quokka weighing 3 pounds is - 2.22 standard deviation(s) below/above the mean. (round to two decimal places as needed.)

Explanation:

Step1: Recall z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value from the data set, $\mu$ is the mean, and $\sigma$ is the standard deviation.

Step2: Calculate z - score for $x = 3$ pounds

Given $\mu = 7$ pounds and $\sigma=1.8$ pounds. Substitute $x = 3$ into the z - score formula: $z=\frac{3 - 7}{1.8}=\frac{-4}{1.8}\approx - 2.22$. A negative z - score means the value is below the mean.

Step3: Calculate z - score for $x = 9$ pounds

Substitute $x = 9$ into the z - score formula: $z=\frac{9 - 7}{1.8}=\frac{2}{1.8}\approx1.11$.

Step4: Determine which is more unusual

The further a value is from the mean in terms of standard deviations, the more unusual it is. Since $| - 2.22|>1.11$, a quokka weighing 3 pounds is more unusual.

Answer:

a) A quokka weighing 3 pounds is - 2.22 standard deviation(s) below the mean.
b) A quokka weighing 3 pounds is more unusual.