Assignment Statistical

Assignment 2

1)a)

 μ = np = 4

σ2 = np(1-p) = 2

np = 2np(1-p)

1 = 2(1-p)

1 = 2- 2p

2p = 1

P = 0.5, n= 8

When probability = 0.5, maximum value of in range of X is 8.

Ans : (C)

b)

c)

d) It should be

“For a normal approximation with continuity correction to a binomial probability,

It suffices to note that (assuming X ~N (n; p) and n is sufficiently large)

P(X < x) = P(Y < x + 0:5); Y ~Bin (np; np (1 􀀀 p))

Anything wrong with this statement?

2) P(tail) = 1/3

Let x denotes the number of “heads” observed in 5 independent tosses of coin.

P( x = 0 and x = 2 and x = 4) = (1/3)^5 + ((1/3)^3)((2/3)^2))

                                                + ((1/3)*(2/3)^4)//

                                                = 0.0864

Mean = 0.0864 x 5  

         = 0.432

[pic 1]

3) let X be the random variable for consumer that feel confident about food safety

X~ B(3.96, 1.36)

1. P( X = 2) = 0.087
2. P ( X = 4) = 0.329
3. P( x >= 2) = 0.107
4. .
5. .
6. This is because np and np(1-p) is smaller than 5. So a Poisson distribution should be needed.

[pic 2]

4) a)  let Y be the random variable for which the number of male mates a queen wasp have.

Y~po(2.7)

1. P( Y = 2) = 0.2449
2. P( Y <= 2) = 0.494
3. P( 1<= Y <=3) = 1 – P(Y = 0) – P(Y > 4)
                          =         1- P(Y=0) – (1- P(Y <=3))
                          = 1 – 0.067 – ( 1 – 0.7141)
                           =0.6471
4. 2.7
5. .
   
   

1. [pic 3]

[pic 4]

1. X ~ N ( 70, 36)

P( 58 < X < 82 ) = 1 – P(X<59 ) – P( X > 81)

                    = 0.954

[pic 5]

6) a. let X be the random variable for the weight of adult green sea urchins

 X ~ N ( 52.0, 17.2^2)

P( 50< X < 60 ) = 0.225

b. P ( x> 40 ) = 0.757

c.

d.

7. a ) 0.08 + 0.08 = 0.16

b. 1. Mean

c) 3

[pic 6]

X ~ B( 21, 6.8^2)