Question

Assume that a procedure yields a binomial distribution with n trials and the probability of success for one trial is p.

n= 1575, p= 2/5

Use the range rule of thumb to find the minimum usual value μ−2σ and the maximum usual value μ+2σ.

μ = _____

σ = _____

μ−2σ = _____

μ+2σ = _____

Answer 1

GIVEN:

A procedure yields a binomial distribution with n trials and the probability of success for one trial is p.

$n=1575$

$p=2/5$

FORMULAS USED:

In binomial distribution,

The formula for mean $(\mu )$ is given by,

$\mu =np$

The formula for standard deviation $(\sigma )$ is given by,

$\sigma =[np(1-p)]^{1/2}$

CALCULATION:

The mean $(\mu )$ is given by,

$\mu =np$

$=1575*(2/5)$

$\mu =630$

The standard deviation $(\sigma )$ is given by,

$\sigma =[np(1-p)]^{1/2}$

  $=[1575*(2/5)(1-(2/5))]^{1/2}$

  $=[630*0.6]^{1/2}$

$\sigma =19.442$

Thus the minimum usual value is given by,

$\mu -2\sigma =630-(2*19.442)$

  $=591.116$$\approx 591$

The maximum usual value is given by,

$\mu +2\sigma =630+(2*19.442)$

  $=668.884$$\approx 669$

Thus the minimum usual value is $591$ and the maximum usual value is $669$.