In: Statistics and Probability
Based on historical data, your team knows what proportion of the company's orders come from Males (and Females). However, your team would like to expand its sales so that men are more proportionately represented. By the end of this year, you think you would like your total proportion of sales to males to be at least 0.45. If you took a simple random sample of 57 current orders, what is the probability that the sample proportion of male customers is greater than 0.45? Note: You should carefully round any intermediate calculations to 4 decimal places to match wamap's approach and calculations. Answer = Incorrect (Enter your answer as a number accurate to 4 decimal places.)
Gender | n | Mean | Variance | Std. dev. | Std. err. | Median | Range | Min | Max | Q1 | Q3 |
---|---|---|---|---|---|---|---|---|---|---|---|
Female | 1404 | 110.26788 | 12292.386 | 110.87103 | 2.958929 | 73.95 | 796.68 | 4 | 800.68 | 32.755 | 150.16 |
Male | 923 | 108.90898 | 12695.331 | 112.67356 | 3.7086953 | 72.15 | 868.45 | 4.04 | 872.49 | 31.62 | 149.11 |
Based on historical data, your team knows what proportion of the company's orders come from Males (and Females). However, your team would like to expand its sales so that men are more proportionately represented. By the end of this year, you think you would like your total proportion of sales to males to be at least 0.45. If you took a simple random sample of 57 current orders, what is the probability that the sample proportion of male customers is greater than 0.45? Note: You should carefully round any intermediate calculations to 4 decimal places to match wamap's approach and calculations. Answer = Incorrect (Enter your answer as a number accurate to 4 decimal places.) Gender n Mean Variance Std. dev. Std. err. Median Range Min Max Q1 Q3 Female 1404 110.26788 12292.386 110.87103 2.958929 73.95 796.68 4 800.68 32.755 150.16 Male 923 108.90898 12695.331 112.67356 3.7086953 72.15 868.45 4.04 872.49 31.62 149.11
The probability that the sample proportion of male customers is greater than 0.45 is 0.00003.
n = 2327
x = 923
p0 = 923/2327
p0 = 0.39665
By applying normal dsitribution:-
z = 4.28
P(z > 4.28) = 0.00003