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dybincka [34]
3 years ago
6

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?
Physics
1 answer:
Sophie [7]3 years ago
4 0

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

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