Question

2) Suppose you have a normal distribution N(10,3), ie. μ=10, σ = 3.
(10 pts) What is the approximate probability that an observation drawn from this
distribution will land between 7 and 13?
(10 pts) 4 and 16?
(10pts) will have a value greater than 16?

Answer 1

Solution :

Given that ,

mean = = 10

standard deviation = = 3

1) P( 7< x < 13) = P((7 - 10)/ 3) < (x - ) / < (13 - 10) / 3) )

= P(-1 < z < 1)

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

= 0.8413 - 0.1587

= 0.6826

Probability = 0.6826

2) P(4 < x < 16) = P((4 - 10)/ 3) < (x - ) / < (16 - 10) / 3) )

= P(-2 < z < 2)

= P(z < 2) - P(z < -2)

= 0.9772 - 0.0228

= 0.9544

Probability = 0.9544

3) P(x > 16) = 1 - P(x < 16)

= 1 - P((x - ) / < (16 - 10) / 3)

= 1 - P(z < 2)

= 1 - 0.9772

= 0.0228

Probability = 0.0228