In: Statistics and Probability
350 babies were born at Neo Hospital in the past 6 months. The average weight for the babies was found to be 6.6 lbs, with a standard deviation of 0.5 lbs. Round to the nearest tenth of a percent.
a. How many babies would you expect to weigh more than 6.5 lbs?
b. How many babies would you expect to weigh more than 7.5 lbs?
c. How many babies would you expect to weigh between 6.5 and 7.5 lbs?
Solution
Back-up Theory
If a random variable X ~ N(µ, σ2), i.e., X has Normal Distribution with mean µ and variance σ2, then,
Z = (X - µ)/σ ~ N(0, 1), i.e., Standard Normal Distribution and hence
P(X ≤ or ≥ t) = P[{(X - µ)/σ} ≤ or ≥ {(t - µ)/σ}] = P[Z ≤ or ≥ {(t - µ)/σ}] .…………...............................................……...…(1)
Probability values for the Standard Normal Variable, Z, can be directly read off from Standard Normal Tables .......(2a)
or can be found using Excel Function: Statistical, NORMSDIST(z) which gives P(Z ≤ z) .....................................…(2b)
Now to work out the solution,
Let X = weight of the babies in lbs. Then, X ~ N(6.6, 0.5) …………..........................................……………………… (3)
µ and σ are given to be 6.6 and 0.5 respectively
Part (a)
Number of babies expected to weigh more than 6.5 lbs
= 350 x proportion of babies expected to weigh more than 6.5 lbs
= 350 x P(X > 6.5)
= 350 x P[Z > {(6.5 – 6.6)/0.5}] [vide (1) and (3)]
= 350 x P(Z > - 0.2)
= 350 x 0.5793 [vide (2b)
= 203 ANSWER 1
Part (b)
Number of babies expected to weigh more than 7.5 lbs
= 350 x proportion of babies expected to weigh more than 7.5 lbs
= 350 x P(X > 7.5)
= 350 x P[Z > {(7.5 – 6.6)/0.5}] [vide (1) and (3)]
= 350 x P(Z > 1.8)
= 350 x 0.0359 [vide (2b)]
= 13 ANSWER 2
Part (c)
Number of babies expected to weigh between 6.5 and 7.5 lbs
= 350 x proportion of babies expected to weigh between 6.5 and 7.5 lbs
= 350 x P(6.5 < X < 7.5)
= 350 x P(0.0.2 < Z < 1.8) [vide steps of Parts (a) and (b)]
= 350 x {P(Z < 1.8) – P(Z < - 0.2)}
= 350 x (0.9641 – 0.4207) [vide (2b)
= 350 x 0.5434
= 190 ANSWER 2
DONE