Question

ELABORATE EXPLANATIONS PLEASEE

2) Consider babies born in the "normal" range of 37–43 weeks gestational age. A paper suggests that a normal distribution with mean μ = 3500 grams and standard deviation σ = 610 grams is a reasonable model for the probability distribution of the continuous numerical variable x = birth weight of a randomly selected full-term baby.

a) What is the probability that the birth weight of a randomly selected fullterm baby exceeds 4000 g?

b) What is the probability that the birth weight of a randomly selected fullterm baby is between 3000 and 4000 g?

c) What is the probability that the birth weight of a randomly selected fullterm baby is either less than 2000 g or greater than 5000 g?

d) What is the probability that the birth weight of a randomly selected fullterm baby exceeds 7 pounds? (Hint: 1 lb = 453.59 g.

e) How would you characterize the most extreme 0.1% of all full-term baby birth weights?

f) If x is a random variable with a normal distribution and a is a numerical constant (a ≠ 0), then y = ax also has a normal distribution. Use this formula to determine the distribution of full-term baby birth weight expressed in pounds (shape, mean, and standard deviation), and then recalculate the probability from part (d).

Answer 1

Solution:

Given in the question

Mean = 3500

Standard deviation = 610

Solution(a)

P(Xbar>4000) =1-p(Xbar<4000)

Z =(4000-3500)/610) = 500/610 = 0.82

From z table we found p-value

P(Xbar>4000) =1- 0.7989 = 0.2011

Solution(b)

P(3000<Xbar<4000) = p(Xbar<4000) -P(Xbar<3000)

Z =(3000-3500)/610 =-0.82

So from Z table we found

P(3000<XBAR<4000) = 0.7989-0.2011 = 0.5978

Solution(c)

P(Xbar<2000) +P(Xbar>5000)

Z =(2000-3500)/610 = -2.46

Z =(5000-3500)/610 = 2.46

So from Z table we found

0.0069 +1-0.9931

0.0069+0.0069 = 0.0138

Solution(d)

P(Xbar>3175.13)

Z =(3175.13-3500)/610 =- 0.53

So from z table

P(Xbar>7pounds)= 1-0.2981 = 0.7019