Question

Having trouble with part C

According to a survey in a country. 34% of adults do not own a credit card. Suppose a simple random sample of 900 adults is obtained. Complete parts (a) through (d) below Click here to view the standard normal distribution table (Rage 1). Click here to view the standard normal distribution table (page 2). C Approximately normal because n SU.USN and np(1-P)2 TU Determine the mean of the sampling distribution of p. HA = 0.34 (Round to two decimal places as needed.) Determine the standard deviation of the sampling distribution of p 04 * 0.016 (Round to three decimal places as needed.) (b) What is the probability that in a random sample of 900 adults, more than 36% do not own a credit card? The probability is 0.1056 (Round to four decimal places as needed) Interpret this probability if 100 different random samples of 900 adults were obtained, one would expect 11 to result in more than 36% not owning a credit card (Round to the nearest Integer as needed.) (C) What is the probability that in a random sample of 900 adults, between 31% and 36% do not own a credit card? The probability is (Round to four decimal places as needed.)

Answer 1

Answer Part c) - 0.8643

Explanation Step 1: Sketch the curve. The probability that 0.31 <x<0.36 is equal to the blue area under the curve. 0.31 0.36 0.28 0. 30 0 .32 0.34 0.36 0.38 0.40 Step 2: Since = 0.34 and o=0.016 we have: 0.31 - 0.34 0.016 X - P(0.31 <x<0.36) = P(0.31 -0.34 < X-4<0.36 -0.34) =P 0.36 -0.34 0.016 Since Z=1" 0.31-0.34 = -1.88 and 0.36 0.34 = 1.25 we have o : 0.016 P(0.31 <x<0.36 )=P( - 1.88 <7<1.25 ) Step 3: Use the standard normal table to conclude that: P( - 1.88 <7<1.25 ) = 0.8643