Question

Suppose that the diameters of golf balls manufactured by a certain company are normally

distributed with mean of 1.96 inches and standard deviation of 0.04 in. A golf ball will be

considered defective if its diameter is less than 1.90 in. or greater than 2.04 in. What is the

percentage of defective balls manufactured by the company?

Answer 1

$\mu=1.96, \sigma=0.04$

To find the percentage of defective balls manufactured by the company golf ball will be consider as defective if diameter less and 1.90 OR greater than 2.02 Within the range of 1.90,2.02 is non defective

$P(1.90<X<2.04)$ is the probability of golf ball is non defective

Standardizing the value

$\begin{aligned} &Z=(X-\mu) / \sigma \\ &Z=(1.90-1.96) / 0.04 \\ &Z=-1.5 \\ &Z=(X-\mu) / \sigma \\ &Z=(2.04-1.96) / 0.04 \\ &Z=2 \\ &\mathrm{P}(1.90<\mathrm{X}<2.04)=\mathrm{P}(-1.5<\mathrm{Z}<2) \\ &=\mathrm{P}(Z<2)-\mathrm{P}(Z<-1.5) \\ &=0.9772-0.0668 \\ &\mathrm{P}(1.90<\mathrm{X}<2.04)=0.9104 \end{aligned}$

Percentage of non defective is $0.9104 * 100 = 91.04\%$

To find the probability of defective balls is $1 - 0.9104 = 0.0896$

Percentage of defective golf ball is $0.0896 * 100 = 8.96\%$