Question

The weights of newborn babies are distributed​ normally, with a mean of approximately 105 oz and a standard deviation of 10 oz. If a newborn baby is selected at​ random, what is the probability that the baby weighs more than 75 ​oz

Answer 1

Answer: 0.9987

Step-by-step explanation:

Given : The weights of newborn babies are distributed​ normally, with a mean of approximately 105 oz and a standard deviation of 10 oz.

i.e. $\mu=105\ oz$  and $\sigma= 10\ oz$

Let x represents the weights of newborn babies.

If a newborn baby is selected at​ random, then the probability that the baby weighs more than 75 ​oz will be :-

$P(x>75)=P(\dfrac{x-\mu}{\sigma}>\dfrac{75-105}{10})\\\\ =P(z>-3 )=P(z<3)\ \ \ [\because\ P(Z>-z)=P(Z<z)]$

$=0.9987$   [using z-value table]

Hence, the required probability = 0.9987