This question is incomplete, the complete question is;
Based on historical data, your team knows what proportion of the company's number of orders come from Indiana residents. However, your team would like to expand its orders beyond the state of Indiana. By the end of this year, you think you would like your total proportion of orders from Indiana to be between 0.66 and 0.73. If you took a simple random sample of 99 current orders, what is the probability that the sample proportion of sales to Indiana residents is between 0.66 and 0.73
Note: You should carefully round any intermediate calculations to 4 decimal places to match wamap's approach and calculations. After you round to 4 places, use THAT rounded value in later steps. You may need an appropriate population proportion (which you actually found in a previous problem).
Some data:
Total population: 2327
Indiana residents: 1845
Other: 482
Frequency table results for Total Sales:
Count = 2327
Total Sales Frequency
0 to 200 1960
200 to 400 303
400 to 600 53
600 to 800 9
800 to 1000 2
Answer:
The probability that the sample proportion of sales to Indiana residents is between 0.66 and 0.73 is 0.0612
Step-by-step explanation:
Given the data in the question;
p" = Indiana residents / Total population
p" = 1845 / 2327
p" = 0.7929
Now, the probability that the sample proportion of sales to Indiana residents is between 0.66 and 0.73 will be;
P( 0.66 < p < 0.73 ) = P{ [x1 - p" / √( p"(1-p")/n )] < p < [x2 - p" / √( p"(1-p")/n )] }
we substitute
= P{ [0.66 - 0.7929 / √( 0.7929(1-0.7929)/99 )] < p < [0.73 - 0.7929 / √( 0.7929(1-0.7929)/99 )] }
= P{ [-0.1329 / 0.0407] < p < [-0.0629 / 0.0407] }
= P{ -3.2654 < z < -1.5455 }
= ⊕( -1.5455) - ⊕(-3.2654)
from the standard normal table;
= 0.0618 - 0.0006
P( 0.66 < p < 0.73 ) = 0.0612
Therefore, the probability that the sample proportion of sales to Indiana residents is between 0.66 and 0.73 is 0.0612
