Question

2. The dietary intake of vitamin C for adults in New York is not normally distributed, but it has a mean of µ = 88.2 mg/day and standard deviation σ = 12.1 mg/day

(a) What is the probability that a randomly selected New York adult’s vitamin C intake is greater than 100 mg/day? Think carefully about this!

(b) What is the probability that a sample of 100 randomly selected New York adults will have a mean greater than 89 mg/day?

(c) What is the probability that a sample of 200 randomly selected New York adults will have a mean between 88 mg/day and 88.5 mg/day?

Answer 1

Solution:

Given that,

mean =   = 88.2

standard deviation = = 12.1

A ) P ( x > 100)

= 1 - P (x < 100 )

= 1 - P ( x -  / ) < ( 100 - 88.2 / 12.1)

= 1 - P ( z < 11.8 / 12.1 )

= 1 - P ( z < 0.98)

Using z table

= 1 - 0.8865

= 0.1135

Probability = 0.1135

B ) n = 100

= 88.2

=  ( /n) = (12.1 / 100 ) = 1.21

P ( > 89 )

= 1 - P (x < 89 )

= 1 - P ( -  / ) < ( 89 - 88.2 / 1.21 )

= 1 - P ( z < 0..8 / 1.21 )

= 1 - P ( z < 0.66)

Using z table

= 1 - 0.7454

= 0.2546

Probability = 0.2546

C ) n = 200

= 88.2

=  ( /n) = (12.1 / 200 ) = 0.8556

P ( 88 < < 88.5 )

P ( 88 - 88.2 / 0.8556 ) < ( -  / ) < ( 88.5 - 88.2 / 0.8556 )

P ( - 0.2/ 0.8556 < z < 0..3 / 0.8556 )

P (- 0.23 < z < 0.35)

  P (Z < 0.35 ) - P ( z < - 0.23)

Using z table

= 0.6368 - 0.4090

= 0.2278

Probability = 0.2278