Question

Anderson Construction Company offers dental insurance to its employees. A recent study by the human resource director shows the annual cost per employee per year followed the normal probability distribution, with a mean of $1,280 and a standard deviation of $420 per year.

1. What percent of the employees cost Anderson more than $1,500 per year for dental insurance?
2. What percent of the employees cost Anderson between $1,500 and $2,000 per year for dental insurance?
3. What percent of the employees did Anderson have to pay more than $1,000 for dental insurance?
4. What was the minimum dollar amount for the top 10% for dental insurance costs?

Answer 1

Solution :

Given that ,

mean = = 1280

standard deviation = = 420

a) P(x >1500 ) = 1 - p( x< 1500 )

=1- p [(x - ) / < (1500-1280) /420 ]

=1- P(z <0.52 )

= 1 - 0.6985 = 0.3015

probability = 0.3015

b)

P( 1500< x < 2000 ) = P[(1500 - 1280)/ 420) < (x - ) /< (2000 - 1280) /420 ) ]

= P( 0.52< z <1.71 )

= P(z < 1.71) - P(z < 0.52 )

Using standard normal table

= 0.9564 - 0.6985 = 0.2579

Probability =0.2579

c)

P(x >1000 ) = 1 - p( x< 1000 )

=1- p [(x - ) / < (1000-1280) /420 ]

=1- P(z < -0.67 )

= 1 - 0.2514= 0.7486

probability = 0.7486

d) Top 10%

P(Z > z ) = 0.10

1- P(z < z) =0.10

P(z < z) = 1-0.10 = 0.90

z = 1.28

Using z-score formula,

x = z * +

x = 1.28*420+1280

x =1817.6

Minimum dollar amount = 1817.6