Respuesta :
Answer:
(a) The approximate distribution of the proportion pf of women who worked = 0.82, the standard distribution = 0.00738
The approximate distribution of the proportion pM of men who worked = 0.88, the standard distribution ≈ 0.01625
(b) pM - pF = 0.06
While pF - pM = -0.06
The difference in the standard deviation ≈ 0.01025
(c) The probability that a higher proportion of women than men worked last year is 0
Step-by-step explanation:
(a) The given information are;
The percentage of college men that were employed last summer = 88%
The percentage of college women that were employed last summer = 82%
The number of college men interviewed in the survey = 400
The number of college women interviewed in the survey = 400
Therefore, given that the proportion of women that worked = 0.82, we have for the binomial distribution;
p = 0.82
q = 1 - 0.82 = 0.18
n = 400
Therefore;
p × n = 0.82 × 400 = 328 > 10
q × n = 0.18 × 400 = 72 > 10
Therefore, the binomial distribution is approximately normal
We have;
[tex]The \ mean = p = 0.82\\\\The \ standard \ deviation, \sigma = \sqrt{ \dfrac{p \times q}{{n} } }= \sqrt{ \dfrac{0.82 \times 0.18}{{400} }} \approx 0.01921\\[/tex]
Therefore, the approximate distribution of the proportion pf of women who worked = 0.82, the standard distribution ≈ 0.01921
Similarly, given that the proportion of male that worked = 0.88, we have for the binomial distribution;
p = 0.88
q = 1 - 0.88 = 0.12
n = 400
Therefore;
p × n = 0.88 × 400 = 352 > 10
q × n = 0.12 × 400 = 42 > 10
Therefore, the binomial distribution is approximately normal
We have;
[tex]The \ mean = p = 0.88\\\\The \ standard \ deviation, \sigma = \sqrt{ \dfrac{p \times q}{{n} } }= \sqrt{ \dfrac{0.88 \times 0.12}{{400} }} \approx 0.01625\\[/tex]
Therefore, the approximate distribution of the proportion pM of men who worked = 0.88, the standard distribution ≈ 0.01625
(b) Given two normal random variables, we have
The distribution of the difference the two normal random variable = A normal random variable
The mean of the difference = The difference of the two means = pM - pF = 0.88 - 0.82 = 0.06
While pF - pM = -0.06
The difference in the standard deviation, giving only the real values, is given as follows;
[tex]The \ difference \ in \ standard \ deviation = \sqrt{ \dfrac{p_1 \times q_1}{{n_1} } -\dfrac{p_2 \times q_2}{{n_2} } }\\\\= \sqrt{\dfrac{0.82 \times 0.18}{{400} }-\dfrac{0.88 \times 0.12}{{400} }} \approx 0.01025\\[/tex]
(c) When there is no difference between the the proportion of men and women that worked last summer, the probability that there is a difference = 0
Therefore, taking 0 as the standard score, we have;
[tex]z = \dfrac{0 - (-0.06)}{0.01025} \approx 5.86[/tex]
Given that the maximum values for a cumulative distribution table is approximately 4, we have that the probability that a higher proportion of women than men worked last year is 0.
