Answer to Question #94981 in Statistics and Probability for sree

Question

The weight of new born baby is normally distributed with a mean of 3.75kg and a
standard deviation of 0.33kg.
i. Find the probability that the weight of new born baby will be not less than 2.8kg.
ii. What is the weight of new born baby that would be in the middle 92% of
the weight?

Answer 1

$P(X\le a)=\frac{1}{0.33\sqrt{2\pi}}\int\limits_{-\infty}^a e^{-\frac{(x-3.75)^2}{2*0.33^2}}dx=$

$=\frac{1}{0.33\sqrt{2\pi}}\int\limits_{-\infty}^{\frac{a-3.75}{0.33}} e^{-\frac{y^2}{2}}d(0.33y+3.75)=$

$=\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{\frac{a-3.75}{0.33}} e^{-\frac{y^2}{2}}dy$ , where $y=\frac{x-3.75}{0.33}$ and so $x=0.33y+3.75$

https://www.math.arizona.edu/~rsims/ma464/standardnormaltable.pdf - it is table of the standard distribution, so $P(X\le a)=Z\bigl(\frac{a-3.75}{0.33}\bigr)$

1)$P(X\ge 2.8)=1-P(X\le 2.8)=1-Z\bigl(\frac{2.8-3.75}{0.33}\bigr)\approx$

$1-Z(-2.88)\approx 1-0.00199=0.99801$

2)From table we see that $Z(1.41)=0.92$

We can find that $\frac{a-3.75}{0.33}=1.41$ while $a=4.2153$ , so $0.92=Z(1.41)=Z\bigl(\frac{4.2153-3.75}{0.33}\bigr)=P(X\le 4.2153)$

Answer:

1)$0.99801$

2)4.2153 kg