Question

8) Scores on an exam have a normal distribution with a mean of 80 and a standard deviation of 12.

a) Find the probability that a person would score above 90.

b) Find the probability that a person would score between 75 and 85.

c) Find the probability that a group of 7 people would have a mean score above 84.

d) Find the score needed to be in the top 10% of the class.

Answer 1

  Solution :

Given that ,

mean = = 80

standard deviation = =12

P(x >90 ) = 1 - P(x <90 )

= 1 - P[(x - ) / < (90-80) / 12]

= 1 - P(z <0.83 )

Using z table,

= 1 -0.7967

=0.2033

b.

Using z table,  

= P(75< x < 85) = P[(75-80) / 12< (x -) / < (85-80) /12 )]

= P(-0.42 < Z <0.42 )

= P(Z <0.42 ) - P(Z <-0.42 )

Using z table,  

= 0.6628 - 0.3372

= 0.3255

c.

n = 7

= 80

= / n = 12 / 7 =4.54

P( > 84) = 1 - P( <84 )

= 1 - P[( - ) / < (84-80) / 4.54]

= 1 - P(z < 0.88)

Using z table,    

= 1 - 0.8106

= 0.1894

d.

P(Z > z) = 10%

= 1 - P(Z < z) = 0.10  

= P(Z < z ) = 1 - 0.10

= P(Z < z ) = 0.9

= P(Z < z ) = 0.9  

z = 1.28

Using z-score formula  

x = z +

x = 1.28*12+80

=95.36