Question

The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 36 and a standard deviation of about 10. Suppose that 16 individuals are randomly chosen. For the group of 16, find the probability that the average percent of fat calories consumed is more than thirty-five (Round to 3 decimal places)

Answer 1

Answer:

Step-by-step explanation:

Mean μ = 36

standard deviation σ = 10

n = 16

standard deviation of the group σ₁ = σ / √n

σ₁ = 10/ √16

= 10 / 4 = 2.5

σ₁  = 2.5

P ( x > 35 ) = 1 - P ( x < 35 )

P ( x < 35 ) = P( z < (35 - 36) / 2.5 ) = P( z < (35 - 36) / 2.5 )

= P( z < 0.4  )

= .3446

P ( x > 35 ) = 1 - .3446

= .6554 .