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In a simple random sample, each individual member of the population has an equal probability of being selected. This means that if Alonzo has a certain probability of being selected, Cheryl should have the same probability, given that they are both members of the same population.
Given that the chance of Alonzo being selected is [tex]\( \frac{1}{200n} \)[/tex], let's denote this probability as [tex]\( P(\text{Alonzo}) \)[/tex].
Since the probability for Cheryl should be the same in a simple random sample, we can write:
[tex]\[ P(\text{Cheryl}) = \frac{1}{200n} \][/tex]
Next, we need to compare this probability with the options provided:
A. [tex]\( \frac{1}{290} \)[/tex]
B. [tex]\( \frac{1}{2900} \)[/tex]
C. [tex]\( \frac{1}{29,000} \)[/tex]
D. [tex]\( \frac{1}{29} \)[/tex]
Clearly, we need to find which of these options matches [tex]\( \frac{1}{200n} \)[/tex]. Let's examine each option one by one:
1. Option A: [tex]\( \frac{1}{290} \)[/tex]
2. Option B: [tex]\( \frac{1}{2900} \)[/tex]
3. Option C: [tex]\( \frac{1}{29,000} \)[/tex]
4. Option D: [tex]\( \frac{1}{29} \)[/tex]
To determine the matching probability, we need to identify how [tex]\( \frac{1}{200n} \)[/tex] can fit one of these forms.
For [tex]\( \frac{1}{200n} \)[/tex] to match any of these options, the denominator [tex]\( 200n \)[/tex] must be equal to 290, 2900, 29,000, or 29 for some value of [tex]\( n \)[/tex].
We check the given fractions:
- If [tex]\( \frac{1}{200n} = \frac{1}{290} \)[/tex], then [tex]\( 200n = 290 \)[/tex], so [tex]\( n = \frac{290}{200} = 1.45 \)[/tex]. This is not a typical integer value for [tex]\( n \)[/tex].
- If [tex]\( \frac{1}{200n} = \frac{1}{2900} \)[/tex], then [tex]\( 200n = 2900 \)[/tex], so [tex]\( n = \frac{2900}{200} = 14.5 \)[/tex]. Again, this is not a typical integer value for [tex]\( n \)[/tex].
- If [tex]\( \frac{1}{200n} = \frac{1}{29000} \)[/tex], then [tex]\( 200n = 29000 \)[/tex], so [tex]\( n = \frac{29000}{200} = 145 \)[/tex]. This is a reasonable integer value for [tex]\( n \)[/tex].
- If [tex]\( \frac{1}{200n} = \frac{1}{29} \)[/tex], then [tex]\( 200n = 29 \)[/tex], so [tex]\( n = \frac{29}{200} = 0.145 \)[/tex]. This is a very impractically small number for [tex]\( n \)[/tex].
Therefore, [tex]\( \frac{1}{29000} = \frac{1}{200n} \)[/tex] works with [tex]\( n = 145 \)[/tex].
Hence, the chance of Cheryl being selected is [tex]\( \frac{1}{29000} \)[/tex]. The correct answer is:
C. [tex]\( \frac{1}{29,000} \)[/tex]
Given that the chance of Alonzo being selected is [tex]\( \frac{1}{200n} \)[/tex], let's denote this probability as [tex]\( P(\text{Alonzo}) \)[/tex].
Since the probability for Cheryl should be the same in a simple random sample, we can write:
[tex]\[ P(\text{Cheryl}) = \frac{1}{200n} \][/tex]
Next, we need to compare this probability with the options provided:
A. [tex]\( \frac{1}{290} \)[/tex]
B. [tex]\( \frac{1}{2900} \)[/tex]
C. [tex]\( \frac{1}{29,000} \)[/tex]
D. [tex]\( \frac{1}{29} \)[/tex]
Clearly, we need to find which of these options matches [tex]\( \frac{1}{200n} \)[/tex]. Let's examine each option one by one:
1. Option A: [tex]\( \frac{1}{290} \)[/tex]
2. Option B: [tex]\( \frac{1}{2900} \)[/tex]
3. Option C: [tex]\( \frac{1}{29,000} \)[/tex]
4. Option D: [tex]\( \frac{1}{29} \)[/tex]
To determine the matching probability, we need to identify how [tex]\( \frac{1}{200n} \)[/tex] can fit one of these forms.
For [tex]\( \frac{1}{200n} \)[/tex] to match any of these options, the denominator [tex]\( 200n \)[/tex] must be equal to 290, 2900, 29,000, or 29 for some value of [tex]\( n \)[/tex].
We check the given fractions:
- If [tex]\( \frac{1}{200n} = \frac{1}{290} \)[/tex], then [tex]\( 200n = 290 \)[/tex], so [tex]\( n = \frac{290}{200} = 1.45 \)[/tex]. This is not a typical integer value for [tex]\( n \)[/tex].
- If [tex]\( \frac{1}{200n} = \frac{1}{2900} \)[/tex], then [tex]\( 200n = 2900 \)[/tex], so [tex]\( n = \frac{2900}{200} = 14.5 \)[/tex]. Again, this is not a typical integer value for [tex]\( n \)[/tex].
- If [tex]\( \frac{1}{200n} = \frac{1}{29000} \)[/tex], then [tex]\( 200n = 29000 \)[/tex], so [tex]\( n = \frac{29000}{200} = 145 \)[/tex]. This is a reasonable integer value for [tex]\( n \)[/tex].
- If [tex]\( \frac{1}{200n} = \frac{1}{29} \)[/tex], then [tex]\( 200n = 29 \)[/tex], so [tex]\( n = \frac{29}{200} = 0.145 \)[/tex]. This is a very impractically small number for [tex]\( n \)[/tex].
Therefore, [tex]\( \frac{1}{29000} = \frac{1}{200n} \)[/tex] works with [tex]\( n = 145 \)[/tex].
Hence, the chance of Cheryl being selected is [tex]\( \frac{1}{29000} \)[/tex]. The correct answer is:
C. [tex]\( \frac{1}{29,000} \)[/tex]
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