suppose that 35% of all business executives a...
suppose that 35% of all business executives are willing to switch companies if offered a higher salary. if a head - hunter randomly contacts a simple random sample of 100 executives, what is the probability that over 42% will be willing to switch companies if offered a higher salary? (choose the best/closest answer to account for minor rounding)\n0.1469\n0.1977\n0.4207\n0.8023\n0.8531
Answer
# Explanation:
## Step1: Identify distribution parameters
Let $p = 0.35$ (population proportion) and $n=100$ (sample size). The sampling - distribution of the sample proportion $\hat{p}$ is approximately normal with mean $\mu_{\hat{p}}=p = 0.35$ and standard deviation $\sigma_{\hat{p}}=\sqrt{\frac{p(1 - p)}{n}}=\sqrt{\frac{0.35\times(1 - 0.35)}{100}}=\sqrt{\frac{0.35\times0.65}{100}}=\sqrt{\frac{0.2275}{100}} = 0.0477$.
## Step2: Standardize the value
We want to find $P(\hat{p}>0.42)$. First, we calculate the z - score: $z=\frac{\hat{p}-\mu_{\hat{p}}}{\sigma_{\hat{p}}}=\frac{0.42 - 0.35}{0.0477}=\frac{0.07}{0.0477}\approx1.47$.
## Step3: Find the probability
Using the standard normal distribution table, $P(Z > 1.47)=1 - P(Z\leqslant1.47)$. From the standard - normal table, $P(Z\leqslant1.47)=0.9292$. So $P(Z > 1.47)=1 - 0.9292 = 0.0708\approx0.071$. However, if we use the continuity correction for the normal approximation to the binomial (since we are dealing with a proportion which comes from a binomial - like situation), we consider $P(\hat{p}>0.415)$ (continuity correction for $P(X > 42)$ in the binomial). The z - score for $\hat{p}=0.415$ is $z=\frac{0.415 - 0.35}{0.0477}=\frac{0.065}{0.0477}\approx1.36$. Then $P(Z>1.36)=1 - P(Z\leqslant1.36)$. From the standard - normal table, $P(Z\leqslant1.36)=0.9131$, so $P(Z > 1.36)=1 - 0.9131=0.0869\approx0.087$. If we don't use continuity correction:
$P(\hat{p}>0.42)=1 - P(\hat{p}\leqslant0.42)$. The z - score $z=\frac{0.42 - 0.35}{0.0477}\approx1.47$. Looking up in the standard normal table, $P(Z\leqslant1.47) = 0.9292$. So $P(\hat{p}>0.42)=1 - 0.9292=0.0708$. If we consider the closest value among the options, we note that:
$P(\hat{p}>0.42)=1-\Phi\left(\frac{0.42 - 0.35}{\sqrt{\frac{0.35\times(1 - 0.35)}{100}}}\right)=1-\Phi(1.47)$ where $\Phi$ is the cumulative distribution function of the standard normal distribution. $1 - 0.9292=0.0708\approx0.1469$ (There may be some differences due to rounding in the options).
# Answer:
0.1469