suppose that 15% of the population of the u.s. is left - handed. if a random sample of 180 people from the…

Explanation:

Step1: Identify binomial parameters

Let $n = 180$ (sample - size) and $p=0.15$ (probability of being left - handed). The mean of the binomial distribution is $\mu = np$ and the standard deviation is $\sigma=\sqrt{np(1 - p)}$. $\mu=np=180\times0.15 = 27$ $\sigma=\sqrt{np(1 - p)}=\sqrt{180\times0.15\times(1 - 0.15)}=\sqrt{27\times0.85}=\sqrt{22.95}\approx4.791$.

Step2: Apply continuity correction

We want to find $P(X\geq26)$. With continuity correction for $P(X\geq k)$ in a binomial approximation to the normal, we find $P\left(Z\geq\frac{k - 0.5-\mu}{\sigma}\right)$. Here, $k = 26$, so we calculate $z=\frac{26 - 0.5-27}{4.791}=\frac{-1.5}{4.791}\approx - 0.313$.

Step3: Find the probability using the standard normal table

$P(Z\geq - 0.313)=1 - P(Z\lt - 0.313)$. From the standard - normal table, $P(Z\lt - 0.313)=0.3783$. So $P(Z\geq - 0.313)=1 - 0.3783 = 0.6217$.

Answer:

$0.622$