Question

(Put in the probability statements where needed)

According to credit card.com, 29% of adults do not own a credit card.

Question 1 ..Suppose a random sample of 500 adults is asked, “Do you own a credit card?” Describe the sampling distribution of the sample proportion of adults who do not own a credit card.

a) Mean (2 decimal places):

b)Standard Deviation (4 decimal places):

Question 2...Show that the distribution of the sample proportion is normal by performing the proper calculations and explaining what the result means.

Question3... What is the probability that in a random sample of 500 adults, less than 32% do not own a credit card? Use phat to represent the random variable.

a)State the probability statement with the random variable:

b) State the z-score, rounded to 2 decimal places:

Question 4 ) What is the probability that in a random sample of 500 adults, more than 30% do not own a credit card? Use phat to represent the random variable.

a)State the probability statement:

b)State the z-score, rounded to 2 decimal places:

Question 5) ..What is the probability that in a random sample of 500 adults, between 25% and 35% do not own a credit card? use phat to represent the random variable.

a)Give the probability statement:

b)State the smaller z-score, rounded to 2 decimal places:

c) State the larger z-score, rounded to 2 decimal places:

Question 6) .. What is the probability that a random sample of 500 adults results in 125 or fewer who do not own a credit card?Use phat to represent the random variable.

a)Give the proportion of interest, rounded to 2 decimal places:

b)Give the probability statement:

c ) State the z-score, rounded to 2 decimal places:

Question 7 ) Would it be unusual for a random sample of 500 adults to result in 125 or fewer who do not own a credit card? Recall that you calculated this probability in the previous question.

Group of answer choices

a)Yes because the probability is less than 0.05

b)Yes because the probability is greater than 0.05

c)No because the probability is less than 0.05

d)No because the probability is greater than 0.05

Answer 1

-0.29 5500 Given Data P=29% la p = 0.29 6) standard deviation of a) Mean (P(1-P) 0 29x(1-0-29) ni 500 -0.0203 - of 2. Normal approximation: np = 500 x 0.29 145 510 N(1-P) 500 (1-0,24) 355 210 Conditions arve Satisfied to use normal approximaticon 3. $ -32% -0132 P(P < 32%-) Plze 5] =P[z< ] 0.24xC-024 - P[z_ 1.4778) - P721:48] a) p<0-32) b) 2 =1:48 ploe31.) 0.906 Hicapp > 30%) =P CÔ7.0.3) 0.32 -0.29 0103 0.0203 500 From z table 0.3-0.29 C.al 0.0209 97. - PC-P) -0.49 0.29X(1-6:*) 500 n From z table 0.25 -0.29 -1.97 P[z 70.149 ] = 1- P[229.149 J-1-0.687 40.312 1 - f (257. < <36t) • 2(1-25 cm<o:35) b) 2: P-P op & -P = 2:96 P[-199< <2.96) + PZ < 2:95] - P[23 +41) = 0.9985-0-24me 12 - 1974 14 2 ) 20.174 0.0203 C) 72 0.35-0-29 0.0203

6. Given n=500 K-125 P(x = 125) 125 500 = 0.25 n a) P = 0.25 b) ¢ [P<0.29] P-P c) za 0.25 -0.29 = -1.97 0.0203 P[z<-1,97] 0.02ug 7. P[*s 125) =P [P<0.25) = 0.0244 yes because the probability is less than 0.05