Question

3)A random variable X~N(0,1). Find the value of q from the Normal Table such that P(X≤q)=0.2624. (Round the result to 2 decimal places.)

4) A random variable X~N(0.11,0. 682 ). Find the probability P(X≤-0.98). (Round the result to 2 decimal places.)

5) A random variable T has t-distribution with degree of freedom 23. Find the t such that P(T≥t)=0.15. (Round the result to 4 decimal places.)

Answer 1

Solution

a ) Given that

Using standard normal table,

P ( X   q ) = 0.2624

p ( X -0.636 ) = 0.2624

q = - 0.64

4 ) Given that,

mean = = 0.11

standard deviation = = 0.682

p ( X -0.98 )

P ( X - /)   (-0.98 - 0.11 / 0.682)

P( z   -1.09 / 0.682 )

P ( z -1.60 )   

Using z table

= 0.0548

Probability = 0.0548

5 ) Given that ,

Using t table

df = 23

P ( T > t ) = 0.15

1 - P ( T <  t0 ) = 0.15

P ( T <  t ) = 1 - 0.15

P ( t < 1.0603 ) = 0.85

t = 1.0603