Question

question 11 a statistician wants to obtain a systematic random sample of size 65 from a population of 7355. what is k? basic funcs trig enter a mathematical expression more.. ing 52. starting with this person, list the numbers

Explanation:

Step1: Recall systematic sampling formula

In systematic sampling, the sampling interval \( k \) is calculated by dividing the population size \( N \) by the sample size \( n \), i.e., \( k=\frac{N}{n} \) (when \( N \) is divisible by \( n \)). Here, \( N = 7355 \) and \( n=65 \).

Step2: Calculate \( k \)

Substitute the values into the formula: \( k=\frac{7355}{65} \)
Calculate the division: \( 7355\div65 = 113.1538\cdots \)? Wait, no, wait, 65*113 = 7345, 7355 - 7345 = 10, so actually, wait, maybe I made a mistake. Wait, 65*113 = 65*(100 + 13)=6500+845 = 7345. Then 7355 - 7345 = 10. Wait, no, maybe the formula is \( k=\lfloor\frac{N}{n} floor \) or when \( N \) is not divisible by \( n \), we still use \( k=\frac{N}{n} \) and round down? Wait, no, let's recalculate 7355 divided by 65. 65*113 = 7345, 7355 - 7345 = 10, so 7355/65 = 113 + 10/65 = 113 + 2/13 ≈ 113.15. But wait, maybe I miscalculated. Wait, 65*113 = 7345, 7355 - 7345 = 10, so 7355/65 = 113.1538... But that can't be. Wait, maybe the population size is 7345? No, the problem says 7355. Wait, no, 65*113 = 7345, 65*114 = 7410, which is more than 7355. Wait, maybe the formula is \( k = \frac{N}{n} \) when \( N \) is divisible by \( n \), otherwise, we take the floor. But 7355 divided by 65: let's do 7355 ÷ 65. 65*100=6500, 7355-6500=855. 65*13=845, 855-845=10. So total is 100+13=113 with a remainder of 10. So \( k = \lfloor\frac{7355}{65} floor = 113 \)? Wait, no, maybe the problem has a typo, or I made a mistake. Wait, 65*113 = 7345, 7355 - 7345 = 10. So if we take \( k = 113 \), then the sample size would be 113*65 = 7345, but we need 65. Wait, maybe the formula is \( k=\frac{N}{n} \) even with a decimal, but in systematic sampling, usually \( k \) is an integer, so we take the integer part. Wait, but let's check again: 7355 ÷ 65. Let's do 65*113 = 7345, 7355 - 7345 = 10, so 7355/65 = 113.1538. But maybe the problem expects us to do 7355 ÷ 65 = 113.15, but that's not an integer. Wait, maybe I miscalculated 65*113. 65*100=6500, 65*13=845, 6500+845=7345. Then 7355-7345=10. So 7355=65*113 +10. So the sampling interval \( k \) is calculated as \( k = \lfloor\frac{N}{n} floor = 113 \), or sometimes we round up, but in systematic sampling, the formula is \( k=\frac{N}{n} \), and if it's not an integer, we can adjust, but in this case, maybe the problem has a mistake, or I made a mistake. Wait, wait, 65*113 = 7345, 7355 - 7345 = 10, so if we take \( k = 113 \), then we can select 65 samples by starting at a random number between 1 and 10, then adding 113 each time. But the question is just to find \( k \), so maybe the answer is 113 (floor) or 114 (ceiling). Wait, let's recalculate 7355 ÷ 65. Let's do 7355 ÷ 65: 65*113 = 7345, 7355 - 7345 = 10, so 7355/65 = 113 + 10/65 = 113 + 2/13 ≈ 113.15. But maybe the problem actually has N=7345? Let's check 7345 ÷ 65: 65*113 = 7345, so k=113. Oh! Maybe a typo in the problem, 7355 should be 7345? Because 65*113=7345. Alternatively, maybe I made a mistake. Wait, 65*113 = 7345, 65*114=7410. So 7355 is between them. But the problem says population of 7355, sample size 65. So \( k = \frac{7355}{65} = 113.1538 \), but in systematic sampling, we usually take \( k = \lfloor\frac{N}{n} floor = 113 \), or sometimes \( k = \lceil\frac{N}{n} ceil = 114 \). But let's check the division again. 65*113 = 7345, 7355 - 7345 = 10, so if we take k=113, then we can have 65 samples (since 113*65=7345, and we have 10 extra, so we can include those 10 in the sample by adjusting the starting point). But the question is just to find k, so maybe the answer is 113 (floor) or 114 (ceiling). Wait, maybe I miscalculated 7355 ÷ 65. Let's do 7355 ÷ 65: 65*100=6500, 7355-6500=855. 855 ÷ 65: 65*13=845, 855-845=10. So 100+13=113, remainder 10. So \( k = 113 \) (since we take the integer part).

Answer:

113