Question

According to government data, 22% of American children under the age of 6 live in households with incomes less than the official poverty level. A study of learning in early childhood chooses an SRS of 300 children. 1. Is it okay to use normal calculations for this problem? Explain (2 pts). 2. Describe the sampling distribution (2 pts). 3. What is the probability that more than 20% of the sample are from poverty households? Draw and shade the appropriate normal curve and find the probability. You must find a z-score and write a conclusion in context (4 pts).

Answer 1

Let denotes number of American children under the age of C live in a households with income less than the official poverty line. under the th an P denotes probability that American children age of live in a household with income official poverty line. .. Po 227 the P 0.22 n: 300 x N Binomial (n 300, P: 0.22) E CX): np 300 x 22 y CX) = npq= 300 x 0.22 x C1-0.22) = 51.48 Hen large & P is small Also np = 66>5, ng - 300 X 0.78: 234>s بند 1 :.X N N C l = np o=npa) ice. x N N C 66, 51.48) OK о ще normal calculations Therefore it this problem, for Normal with mean sampling distribution is variance 51.48. 669 Here n : goo 20 y. of sample : 20 X 300 60 Probability that more than 20% of the sample from parenty hausehold E p Cx>60) X 66 op V 51.48 √51.48 = Pcz 7-0.8362) + co O. 2o33

curve Appropriate Normal - 60 OR -0.8364 0 x-M Z Scox 15 51.48 - .. Required 0.836 2 probability is 0.2093