The distribution of total body protein in healthy adult men is approximately Normal, with mean 12.3 kg and standard deviation 0.1 kg. If you take a random sample of 25 healthy adult men, what is the probability that their average total body protein is between 12.25 and 12.35 kg?

Question

contestada

Respuesta :

wagonbelleville wagonbelleville · Answer 1 · 2019-11-06T09:59:01+01:00

Answer:

[tex]P(12.25\leq x \leq 12.35 ) = 0.9876[/tex]

Explanation:

given,

mean (μ) = 12.3 Kg

standard deviation (σ ) = 0.1

random sample = 25

probability between 12.25 and 12.35 kg

[tex]P(12.25\leq x \leq 12.35 ) = P(\dfrac{12.35-12.3}{\dfrac{0.1}{\sqrt{n}}}\leq z)- P(\dfrac{12.25-12.3}{\dfrac{0.1}{\sqrt{n}}}\leq z)[/tex]

[tex]P(12.25\leq x \leq 12.35 ) = P(\dfrac{12.35-12.3}{\dfrac{0.1}{\sqrt{25}}}\leq z)- P(\dfrac{12.25-12.3}{\dfrac{0.1}{\sqrt{25}}}\leq z)[/tex]

[tex]P(12.25\leq x \leq 12.35 ) = P(\dfrac{12.35-12.3}{\dfrac{0.1}{5}}\leq z)- P(\dfrac{12.25-12.3}{\dfrac{0.1}{5}}\leq z)[/tex]

[tex]P(12.25\leq x \leq 12.35 ) = P(\dfrac{5 (12.35-12.3)}{0.1}\leq z)- P(\dfrac{5(12.25-12.3)}{0.1}\leq z)[/tex]

[tex]P(12.25\leq x \leq 12.35 ) = P(\dfrac{2.5\leq z)- P(-2.5\leq z)[/tex]

using z-table

[tex]P(12.25\leq x \leq 12.35 ) = 0.9938 - 0.0062[/tex]

[tex]P(12.25\leq x \leq 12.35 ) = 0.9876[/tex]