Answer:
0.2078, not unusual
Step-by-step explanation:
Given that we assume that 10% of adults are left-handed.
Let X be the no of left handed persons in the sample of 200 adults
X is binomial because X has two outcomes and each trial is independent of the other
n =200 and p = 0.10
The probability that 16 or fewer are left-handed is
=[tex]P(X\leq 16)\\= F(16)\\=\Sigma _0^{16} 200Cx (0.10)^x (0.9)^{200-x}[/tex]
=0.2075
Since nearly 1/5 are having chance to have left hand this is not unusual.
