Find the circumference of circle with r=28cm

. Suppose (as is roughly true) that 88% of college men and 82% of college women were employed last summer. A sample survey inter

denis23 [38]

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;

$The \ mean = p = 0.82\\\\The \ standard \ deviation, \sigma = \sqrt{ \dfrac{p \times q}{{n} } }= \sqrt{ \dfrac{0.82 \times 0.18}{{400} }} \approx 0.01921\\$

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;

$The \ mean = p = 0.88\\\\The \ standard \ deviation, \sigma = \sqrt{ \dfrac{p \times q}{{n} } }= \sqrt{ \dfrac{0.88 \times 0.12}{{400} }} \approx 0.01625\\$

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;

$The \ difference \ in \ standard \ deviation = \sqrt{ \dfrac{p_1 \times q_1}{{n_1} } -\dfrac{p_2 \times q_2}{{n_2} } }\\\\= \sqrt{\dfrac{0.82 \times 0.18}{{400} }-\dfrac{0.88 \times 0.12}{{400} }} \approx 0.01025\\$

(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;

$z = \dfrac{0 - (-0.06)}{0.01025} \approx 5.86$

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.

3 0

3 years ago

PLEASE HURRY! WILL GIVE 5 STARS! AND THANKS!

solong [7]

Answer:

7/24

Step-by-step explanation:

2/3 - 3/8

2/3 + -3 / 8

16 / 24 + -9 / 24

16 + -9 / 24 = 7/24 (already simplest form, decimal form is 0.291667)

5 0