Question

Some sources report that the weights of​ full-term newborn babies in a certain town have a mean of 9 pounds and a standard deviation of 0.6 pounds and are normally distributed.
a. What is the probability that one newborn baby will have a weight within 0.6 pounds of the meanlong dashthat ​is, between 8.4 and 9.6 ​pounds, or within one standard deviation of the​ mean?
b. What is the probability that the average of nine ​babies' weights will be within 0.6 pounds of the​ mean; will be between 8.4 and 9.6 ​pounds?
c. Explain the difference between​ (a) and​ (b).

Answer 1

Answer:

a) 0.682

b) 0.998

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 9 pounds

Standard Deviation, σ = 0.6 pounds

We are given that the distribution of  weights is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(between 8.4 and 9.6 ​pounds)

$P(8.4 \leq x \leq 9.6) = P(\displaystyle\frac{8.4 - 9}{0.6} \leq z \leq \displaystyle\frac{9.6-9}{0.6}) = P(-1 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < -1)\\= 0.841 - 0.159 = 0.682 = 68.2\%$

$P(8.4 \leq x \leq 9.6) = 68.2\%$

b) P(weight of 9 babies will be between 8.4 and 9.6 ​pounds)

Standard error due to sampling =

$\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{0.6}{\sqrt{9}} = \frac{0.6}{3} = 0.2$

$P(8.4 \leq x \leq 9.6) = P(\displaystyle\frac{8.4 - 9}{0.2} \leq z \leq \displaystyle\frac{9.6-9}{0.2}) = P(-3 \leq z \leq 3)\\\\= P(z \leq 3) - P(z < -3)\\= 0.999 - 0.001 =0.998 = 99.8\%$

$P(8.4 \leq x \leq 9.6) = 99.8\%$

c) In the second part 9 babies were selected from the entire population of babies that caused the major difference in the probabilities for two parts.

Answer 2

Answer:

a) 68.2%

b) 99.8%

c) In the entire population of babies only 9 babies were selected in part b) that causes the major difference between the probabilities of part a) and part b).

Step-by-step explanation:

Given :

Mean, $\mu$ = 9 pounds

Standard deviation, $\sigma$ = 0.6 pounds

And also it is given that it is a normal distribution.

Formula Used :

$z = \dfrac {x - \mu}{\sigma}$

Calculation:

a)

$P(8.4\leq x\leq 9.6) = P(\dfrac {8.4-9}{0.6}\leq z\leq \dfrac{9.6-9}{0.6})= P(-1\leq z\leq 1 )$

$=P(z\leq 1)-P(z<-1)$

= 0.841 - 0.159 = 0.682 = 68.2%

$P(8.4 \leq z\leq 9.6)$ = 68.2%

b)

Due to sampling, standard error is

$\dfrac {\sigma}{\sqrt{n} } = \dfrac {0.6}{\sqrt{9} }=0.2$

$P(8.4\leq x\leq 9.6) = P(\dfrac {8.4-9}{0.2}\leq z\leq \dfrac{9.6-9}{0.2})= P(-3\leq z\leq 3 )$

$=P(z\leq 3)-P(z<-3)=0.999-0.001=0.998$

$P(8.4\leq x\leq 9.6)$ =  99.8%

c)

In the entire population of babies only 9 babies were selected in part b) that causes the major difference between the probabilities of part a) and part b).

For more information, refer the link given below

https://brainly.com/question/11234923?referrer=searchResults