Assume that $14 \%$ of people are left-handed. If 7 people are selected at random, find the probability of each outcome described below.
d) Find the probability that there are exactly 3 lefties in the group.
(Round to four decimal places as needed.)

Step 1 ：Let's assume that 14% of people are left-handed. We are asked to find the probability that exactly 3 out of 7 randomly selected people are left-handed.

Step 2 ：We can use the binomial probability formula to solve this problem. The formula is given by: \( P(X=k) = C(n, k) \cdot p^k \cdot (1-p)^{n-k} \), where \( n \) is the number of trials, \( k \) is the number of successes, \( p \) is the probability of success, and \( C(n, k) \) is the combination of \( n \) items taken \( k \) at a time.

Step 3 ：In this case, \( n = 7 \), \( k = 3 \), and \( p = 0.14 \).

Step 4 ：First, we calculate the combination \( C(n, k) = C(7, 3) \). Using the formula for combinations, we find that \( C(7, 3) = 35 \).

Step 5 ：Next, we substitute \( n = 7 \), \( k = 3 \), \( p = 0.14 \), and \( C(7, 3) = 35 \) into the binomial probability formula to find the probability.

Step 6 ：The calculation is as follows: \( P(X=3) = 35 \cdot (0.14)^3 \cdot (1-0.14)^{7-3} \approx 0.052534663686400015 \).

Step 7 ：Rounding to four decimal places, we find that the probability that exactly 3 out of 7 randomly selected people are left-handed is approximately 0.0525.

Step 8 ：Final Answer: The probability that exactly 3 out of 7 randomly selected people are left-handed is approximately \(\boxed{0.0525}\).