Question

**Use the given values of n and p to find the minimum
usual value µ** **- 2σ** **and the
maximum usual value µ** **+ 2σ.**

n = 1080, p = 0.80

Group of answer choices

Minimum: 850.85; maximum: 877.15

Minimum: 845.41; maximum: 882.59

Minimum: 837.71; maximum: 890.29

Minimum: 890.29; maximum: 837.71

Answer #1

GIven,

From the parameters we say it is a binomial distribution.Having parameters

Then,

Then,

The minimum value is

Maximum value is

Hence,the minimun value is 837.71 and maximum value is 890.29.

Hence, Option C) Minimum:837.71;Maximum:890.29 is correct answer.

Use the given values of n and p to find the minimum usual value
µ - 2σ and the maximum usual value µ + 2σ. n = 95, p = 0.21
Minimum: 27.89; maximum: 12.01 Minimum: 12.01; maximum: 27.89
Minimum: -11.57; maximum: 51.47 Minimum: 15.98; maximum: 23.92

Use the given values of n and p to find the minimum usual value
μ-2σ and the maximum usual value μ+2σ. Round your answer to the
nearest hundredth unless otherwise noted. n = 2112, p = 3/4A)
Minimum: 1478.7; maximum: 1689.3 B) Minimum: 1544.2; maximum:
1623.8 C) Minimum: 1564.1; maximum: 1603.9 D) Minimum: 1623.8;
maximum: 1544.2

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Round to the nearest hundredth unless otherwise noted.
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pequals=0.860.86
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