Respuesta :

fichoh

Complete 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.

Answer: 0.7497

Step-by-step explanation:

Mean = 9

sd = 0.6

n = 4

P(4 babies will be between 8.4 and 9.6 pounds)

P(8.4 ≤ X ≤ 9.6)

Standard error of the sample (S. E) :

S. E = sd/√n

= 0.6 /√4

= 0.6/2

= 0.3

P(8.4 ≤ X ≤ 9.6)= ((P(8.4 - 9) / 0.3) ≤ X ≤ (9.6 - 9)/0.3)

P(8.4 ≤ X ≤ 9.6) = P(-2 ≤ X ≤ 2)

P(Z ≤ 2) - P(Z < - 2)

0.9772 - 0.2275 = 0.7497

Cricetus

The probability will be "0.7497".

Given:

Mean,

  • 9

Standard deviation,

  • 0.6

Number of babies,

  • 4

The standard error of sample will be:

→ [tex]S.E. = \frac{sd}{\sqrt{n} }[/tex]

          [tex]= \frac{0.6}{\sqrt{4} }[/tex]

          [tex]= \frac{0.6}{2}[/tex]

          [tex]= 0.3[/tex]

hence,

→ P(4 babies will be between 8.4 - 9.6 pounds)

[tex]P(8.4 \leq X \leq 9.6) = (\frac{(8.4-9)}{0.3} \leq X \leq \frac{(9.6-9)}{0.3} )[/tex]

                            [tex]= P(-2 \leq X \leq 2)[/tex]

                            [tex]= P(Z \leq 2) - P(Z \leq -2)[/tex]

                            [tex]= 0.9772-0.2275[/tex]

                            [tex]= 0.7497[/tex]

Thus the above response is right.  

Learn more:

https://brainly.com/question/14008215

Otras preguntas