Answer :
Let's walk through solving this problem step-by-step.
Given:
- [tex]\( n = 1585 \)[/tex]
- [tex]\( p = \frac{2}{5} \)[/tex]
The problem involves a binomial distribution, which has important properties like mean and standard deviation.
1. Calculating the Mean ([tex]\(\mu\)[/tex]):
The formula for the mean [tex]\(\mu\)[/tex] of a binomial distribution is given by:
[tex]\[ \mu = n \times p \][/tex]
Plugging in the values:
[tex]\[ \mu = 1585 \times \frac{2}{5} = 1585 \times 0.4 = 634.0 \][/tex]
Therefore, the mean [tex]\(\mu = 634.0\)[/tex].
2. Calculating the Standard Deviation ([tex]\(\sigma\)[/tex]):
The formula for the standard deviation [tex]\(\sigma\)[/tex] of a binomial distribution is given by:
[tex]\[ \sigma = \sqrt{n \times p \times (1 - p)} \][/tex]
Plugging in the values:
[tex]\[ \sigma = \sqrt{1585 \times 0.4 \times (1 - 0.4)} = \sqrt{1585 \times 0.4 \times 0.6} \][/tex]
Calculating the product:
[tex]\[ \sigma = \sqrt{1585 \times 0.24} = \sqrt{380.4} \approx 19.5 \][/tex]
Therefore, the standard deviation [tex]\(\sigma \approx 19.5\)[/tex] (rounded to one decimal place).
3. Using the Range Rule of Thumb:
The range rule of thumb states that usual values lie within the interval:
[tex]\[ [\mu - 2\sigma, \mu + 2\sigma] \][/tex]
Minimum Usual Value ([tex]\(\mu - 2\sigma\)[/tex]):
[tex]\[ \mu - 2\sigma = 634.0 - 2 \times 19.5 = 634.0 - 39.0 = 595.0 \][/tex]
Therefore, the minimum usual value is [tex]\( 595.0 \)[/tex] (rounded to one decimal place).
Maximum Usual Value ([tex]\(\mu + 2\sigma\)[/tex]):
[tex]\[ \mu + 2\sigma = 634.0 + 2 \times 19.5 = 634.0 + 39.0 = 673.0 \][/tex]
Therefore, the maximum usual value is [tex]\( 673.0 \)[/tex] (rounded to one decimal place).
To summarize:
[tex]\[ \mu = 634.0 \][/tex]
[tex]\[ \sigma = 19.5 \][/tex]
[tex]\[ \mu - 2\sigma = 595.0 \][/tex]
[tex]\[ \mu + 2\sigma = 673.0 \][/tex]
Given:
- [tex]\( n = 1585 \)[/tex]
- [tex]\( p = \frac{2}{5} \)[/tex]
The problem involves a binomial distribution, which has important properties like mean and standard deviation.
1. Calculating the Mean ([tex]\(\mu\)[/tex]):
The formula for the mean [tex]\(\mu\)[/tex] of a binomial distribution is given by:
[tex]\[ \mu = n \times p \][/tex]
Plugging in the values:
[tex]\[ \mu = 1585 \times \frac{2}{5} = 1585 \times 0.4 = 634.0 \][/tex]
Therefore, the mean [tex]\(\mu = 634.0\)[/tex].
2. Calculating the Standard Deviation ([tex]\(\sigma\)[/tex]):
The formula for the standard deviation [tex]\(\sigma\)[/tex] of a binomial distribution is given by:
[tex]\[ \sigma = \sqrt{n \times p \times (1 - p)} \][/tex]
Plugging in the values:
[tex]\[ \sigma = \sqrt{1585 \times 0.4 \times (1 - 0.4)} = \sqrt{1585 \times 0.4 \times 0.6} \][/tex]
Calculating the product:
[tex]\[ \sigma = \sqrt{1585 \times 0.24} = \sqrt{380.4} \approx 19.5 \][/tex]
Therefore, the standard deviation [tex]\(\sigma \approx 19.5\)[/tex] (rounded to one decimal place).
3. Using the Range Rule of Thumb:
The range rule of thumb states that usual values lie within the interval:
[tex]\[ [\mu - 2\sigma, \mu + 2\sigma] \][/tex]
Minimum Usual Value ([tex]\(\mu - 2\sigma\)[/tex]):
[tex]\[ \mu - 2\sigma = 634.0 - 2 \times 19.5 = 634.0 - 39.0 = 595.0 \][/tex]
Therefore, the minimum usual value is [tex]\( 595.0 \)[/tex] (rounded to one decimal place).
Maximum Usual Value ([tex]\(\mu + 2\sigma\)[/tex]):
[tex]\[ \mu + 2\sigma = 634.0 + 2 \times 19.5 = 634.0 + 39.0 = 673.0 \][/tex]
Therefore, the maximum usual value is [tex]\( 673.0 \)[/tex] (rounded to one decimal place).
To summarize:
[tex]\[ \mu = 634.0 \][/tex]
[tex]\[ \sigma = 19.5 \][/tex]
[tex]\[ \mu - 2\sigma = 595.0 \][/tex]
[tex]\[ \mu + 2\sigma = 673.0 \][/tex]