Problem 1 The mean credit card debt for a 30 year old male living in NYC is appr

Question

Problem 1

The mean credit card debt for a 30 year old male living in NYC is approximately normally distributed with a mean debt of $8468 and a standard deviation of$1245. (For parts a thru c) What is the probability that a 30 year old male living in NYC selected at random will have:

a) A credit card debt less than $6000?

b) A credit card debt greater that $15000?

c) A credit card debt between $7000 and $10000?

d) What would be the top 10% credit card debt for a 30 year old male livingin NYC?

Problem 2

The mean balance that college students owe on their credit card is $896 with a standard deviation of $117. If all possible random samples of size 169 are taken fromthis population, determine the following:

a) name of the Sampling Distribution

b) mean and standard error of the sampling distribution of the mean (use the correct name and symbol for each)

c) percent of sample means that are less than $880

d) probability that sample means fall between $884 and $900

e) Below which sample mean can we expect to find the lowest 25% of all thesample means?

Problem 3

A recent poll randomly selects 1500 registered voters residing in Nassau County. For this sample 445 respondents state that they are glad that the country has anationwide health plan in place for its citizens.

a) What is the population and parameter that is being discussed?

b) For this parameter identify the point estimate.

c) Determine the estimate of the standard error.

d) Construct a 95% Confidence Interval for the population parameter.

e) Calculate the margin of error.

f) Interpret the meaning of the 95% Confidence Interval for the population parameter.

Explanation / Answer

a) A credit card debt less than $6000?

P(X<6000)=P(Z<6000-8468/1245)

               =P(Z<-1.982)

               =0.02372               .............................Using =NORMSDIST(-1.982) function excel

b) A credit card debt greater that $15000?

P(X>15000)=1- P(Z<15000-8468/1245)

               =1-P(Z<5.24)

               =1-1=0             .............................Using =NORMSDIST(5.24) function excel

c) A credit card debt between $7000 and $10000?

P(7000<X<10000)=P(7000-8468/1245<Z<10000-8468/1245)

               =P(-1.179<Z<1.23)

                =P(Z<1.23)-P(Z<-1.179)

               =0.8907 -0.1191     

               =0.7715

d) What would be the top 10% credit card debt for a 30 year old male livingin NYC?

Here P(Z>z)=0.1

P(Z<z)=1-0.1

P(Z<z)=0.9

z=1.28

x=mean+1.28*standard deviation

=8468+1.28*1245

=10061.6

Hope this will be helpful. Thanks and God Bless you:)

For second quetion repost on Q and A Board.

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.