Assume that a procedure yields a binomial distribution with [tex]\( n \)[/tex] trials and the probability of success for one trial is [tex]\( p \)[/tex]. Use the given values of [tex]\( n \)[/tex] and [tex]\( p \)[/tex] to find the mean [tex]\( \mu \)[/tex] and standard deviation [tex]\( \sigma \)[/tex]. Also, use the range rule of thumb to find the minimum usual value [tex]\( \mu - 2\sigma \)[/tex] and the maximum usual value [tex]\( \mu + 2\sigma \)[/tex].

Given:
[tex]\[ n = 1585, \quad p = \frac{2}{5} \][/tex]

Calculate:
[tex]\[ \mu = \square \][/tex] (Do not round)
[tex]\[ \sigma = \square \][/tex] (Round to one decimal place as needed)
[tex]\[ \mu - 2\sigma = \square \][/tex] (Round to one decimal place as needed)
[tex]\[ \mu + 2\sigma = \square \][/tex] (Round to one decimal place as needed)



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]

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