Question

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?

Answer 1

Answer:

$P(12.25\leq x \leq 12.35 ) = 0.9876$

Explanation:

given,

mean (μ) = 12.3 Kg

standard deviation (σ ) = 0.1

random sample = 25

probability between 12.25 and 12.35 kg

$P(12.25\leq x \leq 12.35 ) = P(\dfrac{12.35-12.3}{\dfrac{0.1}{\sqrt{n}}}\leq z)- P(\dfrac{12.25-12.3}{\dfrac{0.1}{\sqrt{n}}}\leq z)$

$P(12.25\leq x \leq 12.35 ) = P(\dfrac{12.35-12.3}{\dfrac{0.1}{\sqrt{25}}}\leq z)- P(\dfrac{12.25-12.3}{\dfrac{0.1}{\sqrt{25}}}\leq z)$

$P(12.25\leq x \leq 12.35 ) = P(\dfrac{12.35-12.3}{\dfrac{0.1}{5}}\leq z)- P(\dfrac{12.25-12.3}{\dfrac{0.1}{5}}\leq z)$

$P(12.25\leq x \leq 12.35 ) = P(\dfrac{5 (12.35-12.3)}{0.1}\leq z)- P(\dfrac{5(12.25-12.3)}{0.1}\leq z)$

$P(12.25\leq x \leq 12.35 ) = P(\dfrac{2.5\leq z)- P(-2.5\leq z)$

using z-table

$P(12.25\leq x \leq 12.35 ) = 0.9938 - 0.0062$

$P(12.25\leq x \leq 12.35 ) = 0.9876$