Answer to Question #167505 in Statistics and Probability for kienth

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)