Question

Some sources report that the weights of​ full-term newborn babies in a certain town have a mean of

99

pounds and a standard deviation of

0.60.6

pounds and are normally distributed.a. What is the probability that one newborn baby will have a weight within

0.60.6

pounds of the

meanlong dash—that

​is, between

8.48.4

and

9.69.6

​pounds, or within one standard deviation of the​ mean?b. What is the probability that the average of

ninenine

​babies' weights will be within

0.60.6

pounds of the​ mean; will be between

8.48.4

and

9.69.6

​pounds?

c. Explain the difference between​ (a) and​ (b).

a. The probability is

nothing.

​(Round to four decimal places as​ needed.)

Answer 1

Mean weight of newborn babies = = 9 pounds

standard deviaion of newborn babies = = 0.6 pounds

If x is the weight of a random newborn baby then

we have to find

Pr(8.4 pounds < x < 9.6 pounds) = Pr(x < 9.6 pounds ; 9.0 pounds ; 0.6 pounds) - Pr(x < 8.4 pounds ; 9.0 pounds; 0.6 pounds)

Z₂ = (9.6 - 9.0)/0.6 = 1

Z₁ = (8.4 - 9.0)/0.6 = -1

Here checking the Z table for the given z values

Pr(8.4 pounds < x < 9.6 pounds) = Pr(Z < 1) - Pr(Z < -1) = 0.8413 - 0.1587 = 0.6827

So part (a) answer is 0.6827

Now we are taking sample size = n= 9

standard error of sample mean = /sqrt(n) = 0.6/sqrt(9) = 0.2 pounds

Now if is the sample mean of any randomly choosing 9 infants

so here ~ N(9.0 ; 0.2)

we have to find

Pr(8.4 pounds < < 9.6 pounds) = Pr( < 9.6 pounds ; 9.0 pounds ; 0.2 pounds) - Pr( < 8.4 pounds ; 9.0 pounds ; 0.2 pounds)

Z₂ = (9.6 - 9.0)/0.2 = 3

Z₁ = (8.4 - 9.0)/0.2 = -3

Pr(8.4 pounds < < 9.6 pounds) = Pr(Z < 3) - Pr(Z < -3)

Now looing Z table for the given Z values

Pr(8.4 pounds < < 9.6 pounds) = Pr(Z < 3) - Pr(Z < -3) = 0.99865 - 0.00135 = 0.9973

The answer of part (b) is 0.9973.

(c) Here the differnece between part (a) and part(b) is because in part (b) we are talking about sample mean not an individual child.