Question

A normal distribution has a mean of μ = 80 with σ = 12. Find the following probabilities.

(a) p(X > 83)

(b) p(X < 74)

(c) p(X < 92)

(d) p(71 < X < 89)

Answer 1

Solution :

Given that ,

mean = = 80

standard deviation = = 12

P(x >83 ) = 1 - P(x<83 )

= 1 - P[(x -) / < (83 - 80) /12 ]

= 1 - P(z <0.25 )

Using z table

= 1 - 0.5987

= 0.4013

probability= 0.4013

(B)

P(X< 74) = P[(X- ) / < (74 - 80) /12 ]

= P(z <-0.5 )

Using z table

= 0.3085

probability=0.3085

(C)

P(X< 92) = P[(X- ) / < (92 - 80) /12 ]

= P(z <1)

Using z table

= 0.8413

probability=0.8413

(D)

P(71< x < 89) = P[(71 -80) / 12< (x - ) / < (89 -80) /12 )]

= P( -0.75< Z <0.75 )

= P(Z <0.75 ) - P(Z < -0.75)

Using z table   

= 0.7734 -0.2266

   probability= 0.5468