Question

1. Suppose that a random sample of size 64 is to be selected from a population with mean 40 and standard deviation 5.

a. What is the mean of the ¯xx¯ sampling distribution?

b. What is the standard deviation of the ¯xx¯ sampling distribution?

c. What is the approximate probability that ¯xx¯ will be within 0.5 of the population mean μμ?

d. What is the approximate probability that ¯xx¯ will differ from μμ by more than 0.7?

2. A Food Marketing Institute found that 45% of households spend more than $125 a week on groceries. Assume the population proportion is 0.45 and a simple random sample of 190 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is between 0.23 and 0.49?

Answer =  (Enter your answer as a number accurate to 4 decimal places.)

3. Based on historical data, your manager believes that 29% of the company's orders come from first-time customers. A random sample of 193 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is less than 0.31?

Answer =  (Enter your answer as a number accurate to 4 decimal places.)

Answer 1

1)

a) mean of the ¯x 40

b) standard deviation of the ¯x =5/sqrt(64)=0.625

c)for normal distribution z score =(X-μ)/σx

probability that ¯xx¯ will be within 0.5 of the population mean μ:

probability =

P(39.5<X<40.5)

=

P(-0.8<Z<0.8)=

0.7881-0.2119=

0.5762

d)approximate probability that ¯xx¯ will differ from μμ by more than 0.7 :

P(X>40.7)+P(X<39.3)=1-P(39.3<X<40.7)=1-P(-1.12<X<1.12)=1-(0.8686-0.1314)=0.2628

2)

for normal distribution z score =(p̂-p)/σp
here population proportion=     p=	0.450
sample size       =n=	190
std error of proportion=σp=√(p*(1-p)/n)=	0.0361

probability =

P(0.23<X<0.49)

=

P(-6.1<Z<1.11)=

0.8665-0=

0.8665

3)

for normal distribution z score =(p̂-p)/σp
here population proportion=     p=	0.290
sample size       =n=	193
std error of proportion=σp=√(p*(1-p)/n)=	0.0327

probability =

P(X<0.31)

=

P(Z<0.61)=

0.7291