Question

Use the following information for problems 4-6.The mean weight of a domestic house cat is 8.9 pounds, with a standard deviation of 1.1 pounds.The distribution of domestic house cat weights can reasonably be approximated by a normal distribution.What percentage of domestic house cats weigh less than 7.8 pounds?What percentage of domestic house cats weigh more than 11.1 pounds?A domestic house cat weighing 9 pounds is in what percentile?Use the following information for problems 7-9.An article in a health magazine suggested getting a dog in order to increase time spent walking (for exercise). A researcher for the article found that the distribution of time spent walking by dog-owners was approximately normally distributed with a mean of 38 minutes per day. She also found that 84% of dog-owners spend less than 45 minutes walking per day.Find the standard deviation (in minutes) for daily walking time among dog-owners.Construct a normal distribution curve that displays all relevant data for this scenario.How long would a dog-owner need to walk in a day to be ranked in the 99th percentile?Using any research tools at your disposal, find 3 real-world examples of variables that are, in fact, normally distributed.

Answer 1

In order to solve this question, we need to use the z-score z of value x, belonging to a normal distribution with mean μ and standard deviation σ.

z is defined as:

`z=(x-\mu)/(\sigma)`

In this problem, we have, in pounds:

μ = 8.9

σ = 1.1

PART 1

So, the percentage of domestic house cats that weigh less than 7.8 pounds is:

`P(x<7.8)=P(z<(7.8-8.9)/(1.1))=P(z<-1)`

And from a z-score table, we have:

`P(z<-1)=0.15866`

Therefore, the percentage of domestic house cats that weigh less than 7.8 pounds is approximately 0.1587 or 15.87%.

PART 2

The percentage of domestic house cats that weigh more than 11.1 pounds is 1 minus the percentage of them that weigh less than that:

`\begin{gathered} P(x>11.1)=1-P(x<11.1) \\ \\ P(x>11.1)=1-P(z<(11.1-8.9)/(1.1))=1-P(z<2) \\ \\ P(x>11.1)=1-0.97725=0.02275 \end{gathered}`

Therefore, the percentage of domestic house cats that weigh more than 11.1 pounds is approximately 0.0228 or 2.28%.

PART 3

The domestic house cat weighing 9 pounds is in percentile P(x<9):

`P(x<9)=P(z<(9-8.9)/(1.1))=P(z<0.09091)=0.53622\cong0.54`

Therefore, a domestic house cat weighing 9 pounds is in the 54th percentile.