I only need Parts e) and f) please show work and how to do the problem. 698 gram
ID: 3059750 • Letter: I
Question
I only need Parts e) and f) please show work and how to do the problem.
698 grams is a reasonable Consider babies born in the normal range of 37-43 weeks gestational age. A paper suggests that a normal distribution with mean = 3500 grams and standard deviation 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 full-term baby exceeds 4000 g? (Round your answer to four decimal places.) 2358 (b) What is the probability that the birth weight of a randomly selected full-term baby is between 3000 and 4000 g? (Round your answer to four decimal places.) 0.5284 (c) What is the probability that the birth weight of a randomly selected full-term baby is either less than 2000 g or greater than 5000 g? (Round your answer to four decimal places.) 0.0316 (d) What is the probability that the birth weight of a randomly selected full-term baby exceeds 7 pounds? (Hint: 1 lb 453.59 g. Round your answer to four decimal places.) 0.6808 (e) How would you characterize the most extreme 0.1% of all full-term baby birth weights? (Round your answers to the nearest whole number.) The most extreme 0.1% of birth weights consist of those greater than 5945 grams and those less than 1030 X grams. (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). (Round your answer to four decimal places.)Explanation / Answer
Solution:
Let X denotes the birth weight of babies
Given that birth weight is normal distributed with mean 3500 g and standard deviation 698g, That is X — N(3500,698)
(a) The probability that the birth weight of a randomly selected baby of this type exceeds 4000 grams is,
P(x > 4000) = P(Z > 4000-3500)/698 ) =P(Z > 0.72)
= 0.2358
(b) Probability that the birth weight of a randomly selected baby between 3000 and 4000 grams is,
P(3000 < x < 4000) = P(3000 -3500/698 < Z < 4000-3500/698)
= P(-0.72 < Z < 0.72)
= 0.5284
(c) The probability that the birth weight of a randomly selected baby either less than 2000 grams or greater than 5000 grams is,
P(x < 2000 or x > 5000) = P(x < 2000) + P(x > 5000)
= P(z < 2000 -3500/698 ) + P(z > 5000 -3500/698)
= P(Z < -2.15) + P(Z > 2.15)
= 0.0158+ 0.0158
= 0.0316
(d) The probability that the birth weight of a randomly selected baby exceeds 7 pounds is,
Here convert lb in terms of grams. It is known that 1 lb (= 453.59 grams than 7 lbs = 3175.13 grams.
P(x > 3175.13) =P(Z > 3175.13 -3500/698)
= P(Z > -0.47)
= 0.6808
(e) The most extreme 0.1% of all birth weights is,
P(-c < Z < c) = 0.999
=> P(Z -c) +P(Z c) = 0.001
=> 2P(Z -c) = 0.001
=> P(Z < -c) = 0.0005
=> c = 3.29
The highest extreme value is,
X-/ = 3.29
=> X = 3.29 x 698 + 3500
X = 5796.42 5796
The smallest extreme value is,
X-µ/ = -3 29
=> X = -3.29 x 698+3500
X = 1203.58
X = 1204
Thus the most extreme 0.1% of all birth weights is,
X <1204 or X > 5796
(f) It is known that 1 lb = 453.59 gm, then
Y = X/453.59
The expected value of Y is,
E(Y) = E(X/453.59) = 7 .566 Lb
The standard deviation of Y is,
= Var(Y) = Var(X/453.59)
= 1/453.59Var(X)
= x/453.59
= 698/453.59
= 1.53 lb