Question

Recall that 40% of Americans have an associates degree or above. A group of 80 people have applied for a job (which does not list any educational restrictions) and you will be reviewing their applications What sampling distribution model would you use to model the probability of seeing various sample proportions of degree-holders in a group this size? a. b. Use your model to find the probability that 50% or more of your applicant pool will have degrees?

Answer 1

a) Let denotes the sample proportion of Americans that have an associate degree or above.

a) Let p denotes the sample proportion of Americans that have an associate degree or above. p tilde Normal(0.4, 0.4 * (1 - 0.4)/80) or p tilde Normal(0.4, 0.0547723^2) P(p greaterthanorequalto 0.5) = 1 - P(p lessthanoreqaualto 0.5) = 1 - P(p - 0.4/squareroot 0.4 * (1 - 0.4)/80 lessthanoreqaualto 0.5 - 0.4/squareroot 0.4 * (1 - 0.4)/80) b) Rightarrow P(p greaterthanorequalto 0.5) = 1 - P(Z lessthanoreqaualto 0.5 - 0.4/0.0547723) Rightarrow P(p greaterthanorequalto 0.5) = 1 - P(Z lessthanoreqaualto 0.1/0.0547723) Rightarrow P(p greaterthanorequalto 0.5) = 1 - P(Z lessthanoreqaualto 1.82574) Rightarrow P(p greaterthanorequalto 0.5) = 1 - phi(1.82574) Rightarrow P(p greaterthanorequalto 0.5) = 1 - 0.966055 = 0.033945 Rightarrow P(p greaterthanorequalto 0.5) = 0.033945

$\hat{p} \sim Normal(0.4,\frac{0.4*(1-0.4)}{80})$ or $\hat{p} \sim Normal(0.4,0.0547723^2)$

b) $P(\hat{p} \geq 0.5) = 1- P(\hat{p} \leq 0.5) =1- P(\frac{\hat{p}-0.4}{\sqrt{\frac{0.4*(1-0.4)}{80}}} \leq \frac{0.5-0.4}{\sqrt{\frac{0.4*(1-0.4)}{80}}})$

$\Rightarrow P(\hat{p} \geq 0.5) = 1- P(Z \leq \frac{0.5-0.4}{0.0547723})$

$\Rightarrow P(\hat{p} \geq 0.5) = 1- P(Z \leq \frac{0.1}{0.0547723})$

$\Rightarrow P(\hat{p} \geq 0.5) = 1- P(Z \leq 1.82574)$

$\Rightarrow P(\hat{p} \geq 0.5) = 1- \Phi(1.82574)$

$\Rightarrow P(\hat{p} \geq 0.5) = 1-0.966055 = 0.033945$

$\Rightarrow P(\hat{p} \geq 0.5) =0.033945$