Question

Assume that a procedure yields a binomial distribution with n trials and the probability of success for one trial is p.Use the given values of n and p to find the mean μ and standard deviation σ.Use the range rule of thumb to find the minimum usual value μ−2σ and the maximum usual value μ+2σ.n=1415, p=3/5 μ

Answer 1

Answer:
`\mu=849`

`\sigma=18.43`

Minimum usual value
`=\mu-2\sigma=849-2(18.43)=849-36.86=812.14`

Maximum usual value
`=\mu+2\sigma=849+2(18.43)=849+36.86=885.86`

Explanation:

The Normal approximation can be used to binomial distribution, when n is large and p is near to 0.5.

`X\sim N(np,√(np(1-p)))`

here, Mean =
`\mu=np`

Standard deviation :
`\sigma=√(np(1-p))`

Given :
`n= 1415,\ p=(3)/(5)`

Then,
`\mu=1415*(3)/(5)=849`

`\sigma=\sqrt{1415((3)/(5))(1-(3)/(5))}\\\\=√(339.6)\approx18.43`

Range rule of thumb : It says that the range is approximately four times the standard deviation.

Minimum usual value
`=\mu-2\sigma=849-2(18.43)=849-36.86=812.14`

Maximum usual value
`=\mu+2\sigma=849+2(18.43)=849+36.86=885.86`