Question

The weights of 1,000 children, in average, is 51kg with standard deviation of 11kg. Suppose the weights are normally distributed, how many children weigh between 31kg and 69kg

Answer 1

The weights of the $1000$ children are normally distributed with mean $51$ kg and standard deviation $11$ kg.

Let $X$ be a random variable denotes the weights of the children.

Then $\mu =51$ and $\sigma =11$ .

Let $Z= \frac{X-\mu }{\sigma}$ . Then $Z=\frac{X-51}{11}$.

Here we have to find $P(31<X<69)$ .

Now $P(31<X<69)=P(\frac{31-51}{11}<Z<\frac{69-51}{11})$

$=P(-1.82<Z<1.64)$

$=P(0<Z<1.64)+P(0<Z<1.82)$

$=$ $(0.4495+0.4656)$

$=0.9151$

Therefore number of children's weight lie between $31$ kg and $69$ kg are $=(1000×0.9151)=915$ (approximately)