Question

Rounded to 4 decimal, what is the standard deviation of a binomila distribution that uses 216 trials with a success probability of 0.17?

Rounded to 4 decimal, what is the standard deviation of a binomila distribution that uses 23 trials with a success probability of 0.85?

Rounded to 4 decimal, what is the standard deviation of a binomila distribution that uses 31 trials with a success probability of 0.19?

Answer 1

Solution :

Given that ,

mean =   =

standard deviation = =   

P(X< ) = P[(X- ) / < () / ]

= P(z < )

Using z table

=   

Solution :

Given that ,

mean =   =

standard deviation = =     

P(x  )

= P[(x - ) /   () / ]

= P(z )

Using z table,

=

Solution :

Given that ,

mean =   =

standard deviation = =

n =

=

=  / n = /

P( < ) = P[( - ) / < () / ]

= P(z < )

Using z table  

=   

probability=   

Solution

Given that,

p =

1 - p =

n =

= p =

=  [p ( 1 - p ) / n] =   [(0.) / ] =

P( < ) =

= P[( - ) / < () / ]

= P(z < )

Using z table,   

=

probability

Solution :

Given that,

p = 0.17

q = 1 - p =1-0.17=0.83

(A)n = 216

Using binomial distribution,

Standard deviation = = n * p * q =  216*0.17*0.83=5.5207

(B)

p = 0.85

q = 1 - p =1-0.85=0.15

n = 23

Using binomial distribution,

Standard deviation = = n * p * q =  23*0.85*0.15=1.7125

(C)

p = 0.19

q = 1 - p =1-0.19=0.81

n = 31

Using binomial distribution,

Standard deviation = = n * p * q =  31*0.19*0.81=2.1842