Step 1 :Given that the sample size (n) is 200 and the probability of success (p), which is the probability of an adult not owning a credit card, is 0.25.
Step 2 :Calculate the mean of the binomial distribution using the formula \(np\), which gives \(200 \times 0.25 = 50.0\).
Step 3 :Calculate the standard deviation of the binomial distribution using the formula \(\sqrt{np(1-p)}\), which gives \(\sqrt{200 \times 0.25 \times (1-0.25)} = 6.123724356957945\).
Step 4 :Standardize the value 0.28 to find the corresponding z-score using the formula \(\frac{0.28 - p}{\sqrt{\frac{p(1-p)}{n}}}\), which gives \(\frac{0.28 - 0.25}{\sqrt{\frac{0.25(1-0.25)}{200}}} = 0.9797958971132724\).
Step 5 :Find the probability that the z-score is greater than this value, which is approximately 0.1636.
Step 6 :Thus, the probability that in a random sample of 200 adults, more than 28% do not own a credit card is approximately \(\boxed{0.1636}\).