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

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

Step-by-step 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((12.35-12.3)/((0.1)/(√(n)))\leq z)- P((12.25-12.3)/((0.1)/(√(n)))\leq z)`

`P(12.25\leq x \leq 12.35 ) = P((12.35-12.3)/((0.1)/(√(25)))\leq z)- P((12.25-12.3)/((0.1)/(√(25)))\leq z)`

`P(12.25\leq x \leq 12.35 ) = P((12.35-12.3)/((0.1)/(5))\leq z)- P((12.25-12.3)/((0.1)/(5))\leq z)`

`P(12.25\leq x \leq 12.35 ) = P((5 (12.35-12.3))/(0.1)\leq z)- P((5(12.25-12.3))/(0.1)\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`