Answer to Question #118632 in Statistics and Probability for Michael

Question

The weights of 100 remote control cars at a competition are approximately normally distributed. The average weight is 3.2 kg , with a standard deviation of 0.4 kg

a. How many remote control cars would be disqualified if it were against the rules to have a car with a weight of more than 4 kg or less than 2.4 kg ?
b. A car is said to be in the 90th percentile. How much does it weigh?

Answer 1

Let $X=$ the weight of the car: $X\sim N(\mu, \sigma^2).$ Then $Z=\dfrac{X-\mu}{\sigma}\sim N(0, 1)$

Given $\mu=3.2\ kg, \sigma=0.4\ kg.$

a.

$$P(2.4<X<4)=P(X<4)-P(X\leq 2.4)=$$

$$=P(Z<{4-3.2\over 0.4})-P(Z\leq {2.4-3.2\over 0.4})=$$

$$=P(Z<2)-P(Z\leq-2)\approx0.977250-0.022750=$$

$$=0.9545 (95.45 \ \%)$$

$0.9545\cdot100=95.45\approx 95$

5 remote control cars would be disqualified.

b. 90^th percentile means that 90% of the values are below $z^*$

$$P(Z<z^*)=0.9=>z^*\approx1.281552$$

$$z^*={x^*-\mu\over \sigma}=>x^*=z^*\sigma+\mu$$

$$x^*=1.281552(0.4)+3.2=3.7(kg)$$

A car weighs 3.7 kg.