Question

Draw a normal curve and shade it appropriately for the question below It is known that 75% of American adults have a credit card. If we take a random sample of American adults, what is the probability that more than 78% of the people sampled have credit cards? A pollster claims that the mean amount spent on Thanksgiving gifts by an American family is $345 with a standard deviation of S150. What is the probability that the mean amount spent on Thanksgiving gifts for a sample of 400 families is between $340 and 365?

Answer 1

Solution-5:

mean=p^=0.75

n=425

standard devaition=sqrt(p*(1-p)/n)

=sqrt(0.75*(1-0.75)/425)

=0.0210042

Need to find

P(p^>0.78)

=P(Z>0,78-0.75/0.0210042)

=P(Z> 1.4283)

=1-P(Z<1.4283)

=1-0.9234

=0.0766

ANSWER:

0.0766

Normal Curve, mean = 0.75, SD = 0.02 Shaded Area = 0.0766 density 10 0 0.75 0.78

Solution2:

X~N(345,150)

xbar=345

s=sigma/sqrt(n)=150/sqrt(400)=150/20=7.5

P(340<xbar<365)

=P(340-345/7.5<xbar<365-345/7.5)

=P(-0.6667<Z<2.6667)

=P(Z<2.6667)-P(Z<-0.6667)

=0.9962-0.2525

=0.7437

Normal Curve, mean = 345, SD = 7.5 Shaded Area = 0.7437 0.05 density 0.00 340 365