thanks to social media, the happiest creature on earth, an always smiling australian marsupial called a…

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 described by 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 the mean. (round to two decimal places as needed.) b) a quokka weighing 9 pounds is standard deviation(s) the mean. the quokka weighing pounds is more unusual because its weight is a number of standard deviations from the mean weight. (round to two decimal places as needed.)

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 described by 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 the mean. (round to two decimal places as needed.) b) a quokka weighing 9 pounds is standard deviation(s) the mean. the quokka weighing pounds is more unusual because its weight is a number of standard deviations from the mean weight. (round to two decimal places as needed.)

Answer

Explanation:

Step1: Calculate z - score for 3 - pound quokka

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value, $\mu$ is the mean, and $\sigma$ is the standard deviation. Given $\mu = 7$, $\sigma=1.8$, and $x = 3$. Then $z=\frac{3 - 7}{1.8}=\frac{-4}{1.8}\approx - 2.22$.

Step2: Calculate z - score for 9 - pound quokka

Using the same formula with $x = 9$, $\mu = 7$, and $\sigma=1.8$. Then $z=\frac{9 - 7}{1.8}=\frac{2}{1.8}\approx1.11$.

Step3: Compare z - scores

The more unusual value is the one with the larger absolute - value of the z - score. Since $| - 2.22|=2.22$ and $|1.11| = 1.11$, and $2.22>1.11$.

Answer:

a) A quokka weighing 3 pounds is 2.22 standard deviation(s) below the mean. b) A quokka weighing 9 pounds is 1.11 standard deviation(s) above the mean. The quokka weighing 3 pounds is more unusual because its weight is a larger number of standard deviations from the mean weight.