Answer: a. 0.9876
Step-by-step explanation:
Given : The distribution of total body protein in healthy adult men is approximately Normal with mean 12.3 kg and standard deviation 0.1 kg.
[tex]\mu=12.3[/tex] and [tex]\sigma=0.1[/tex]
sample size : n= 25
Let x denotes the total body protein in healthy adult men in the sample.
Then, the probability that their average total body protein is between 12.25 and 12.35 kg will be :-
[tex]P(12.25<x<12.35)=P(\dfrac{12.25-12.3}{\dfrac{0.1}{\sqrt{25}}}<\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}<\dfrac{12.35-12.3}{\dfrac{0.1}{\sqrt{25}}})\\\\=P(\dfrac{-0.05}{\dfrac{0.1}{5}}<\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}<\dfrac{0.05}{\dfrac{0.1}{5}})\\\\=P(-2.5<z<2.5)\ \ [\because\ z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=P(z<2.5)-P(z<-2.5)\\\\=P(z<2.5)-(1-P(z<2.5))\ \ [\because\ P(Z<-z)=1-P(Z<z)]\\\\=2P(z<2.5)-1\\\\=2(0.9938)-1\ \ [\text{By using z-table}]\\\\=1.9876-1 =0.9876[/tex]
∴ The correct answer is a. 0.9876
Average total body protein is between 12.25 and 12.35 kg, the probability will be "0.9876".
According to the question,
Mean, μ = 12.3
Standard deviation, σ = 0.1
Sample size, n = 25
The probability be:
→ P (12.25 < x < 12.35) = [tex]P(\frac{12.25-12.3}{\frac{0.1}{\sqrt{25} } }< \frac{x- \mu}{\frac{\sigma}{\sqrt{n} } } < \frac{12.35-12.3}{\frac{0.1}{\sqrt{25} } } )[/tex]
= [tex]P(\frac{-0.05}{\frac{0.1}{5} } < \frac{x - \mu}{\frac{\sigma}{\sqrt{n} } } < \frac{0.05}{\frac{0.1}{5} } )[/tex]
= [tex]P(-2.5 = [tex]P(z<2.5) -P(z<-2.5)[/tex]
= [tex]2P(z< 2.5)-1[/tex]
By using the z-table, we get
= [tex]2(0.9938)-1[/tex]
= [tex]1.9876-1[/tex]
= [tex]0.9876[/tex]
Thus the above answer i.e., "option a" is correct.
Find out more information about probability here:
https://brainly.com/question/24756209
