Answer:
a) Mean of sample proportion = 0.16
Mean of random sample that are minority member applications = 336
b) Standard deviation of p = 0.008
In number terms, it is 16.8 applications.
c) Probability that there will be 15% or fewer minority member applications in a random sample of 2100 drawn from all applications = 0.1057
Step-by-step explanation:
Complete Question
A minority representation group accuses a major bank of racial discrimination in its recent hires for financial analysts. Exactly 16% of all applications were from minority members, and exactly 15% of the 2100 open positions were filled by members of the minority.
a. Find the mean of p, where p is the proportion of minority member applications in a random sample of 2100 that is drawn from all applications.
b. Find the standard deviation of p.
c. Compute an approximation for P (p ≤ 0.15), which is the probability that there will be 15% or fewer minority member applications in a random sample of 2100 drawn from all applications. Round your answer to four decimal places.
Solution
a) If the sample is a random sample, he proportion of the applications that will be minority member applications is still going to be approximately the same as the population proportion of the minority member applications.
Mean proportion = p = 0.16
The mean of the sample is then given with is given as
Mean = np
where n = sample size = 2100
p = mean proportion = 16% = 0.16
mean = 0.16 × 2100 = 336
b) The standard deviation of p is given as
σ = √[p(1-p)/n]
= √[0.16×0.84/2100] = 0.008
In real number terms,
= 0.008×2100 = 16.8 applications
c) Probability that there will be 15% or fewer minority member applications in a random sample of 2100 drawn from all applications
P(p ≤ 0.15)
We will use the normal approximation to the binomial distribution since, np and np(1-p) are both greater than 10.
Mean = 0.16
standard deviation = 0.008
To find P(p ≤ 0.15)
We first need to normalize or standardize 0.15
The standardized score for any value is the value minus the mean then divided by the standard deviation.
z = (x - μ)/σ = (0.15 - 0.16)/0.008 = -1.25
To determine the required probability
We'll use data from the normal distribution table for these probabilities
P(p ≤ 0.15) = P(z ≤ -1.25) = 0.10565 = 0.1057 to 4 d.p.
Hope this Helps!!!
