According to a survey in a country, $28 \%$ of adults do not own a credit card. Suppose a simple random sample of 700 adults is obtained. Complete parts (a) through (d) below. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). The probability is 0.1192 (Round to four decimal places as needed.) Interpret this probability. If 100 different random samples of 700 adults were obtained, one would expect 12 to result in more than $30 \%$ not owning a credit card. (Round to the nearest integer as needed.) (c) What is the probability that in a random sample of 700 adults, between $26 \%$ and $30 \%$ do not own a credit card? The probability is $\square$. (Round to four decimal places as needed.)

Step 1 ：First, we need to calculate the standard deviation of the proportion. The proportion (p) is 0.28 and the sample size (n) is 700. The standard deviation of the proportion (\(\sigma\)) is calculated as \(\sqrt{p \times (1 - p) / n}\), which gives us \(\sigma = 0.01697056274847714\).

Step 2 ：Next, we calculate the z-scores for 26% and 30%. The z-score is calculated as \((x - p) / \sigma\), where x is the value for which we want to find the z-score. For 26%, the z-score (z1) is \((0.26 - p) / \sigma = -1.1785113019775804\). For 30%, the z-score (z2) is \((0.30 - p) / \sigma = 1.178511301977577\).

Step 3 ：Then, we calculate the probabilities corresponding to these z-scores using the standard normal distribution. The probability corresponding to z1 is 0.1192964146582175 and the probability corresponding to z2 is 0.880703585341782.

Step 4 ：Finally, we find the difference between these probabilities to get the probability that in a random sample of 700 adults, between 26% and 30% do not own a credit card. The probability is \(p2 - p1 = 0.7614071706835644\).

Step 5 ：The final answer is \(\boxed{0.7614}\), which means there is approximately a 76.14% chance that in a random sample of 700 adults, between 26% and 30% do not own a credit card.