Question

he length of similar components produced by a company are approximated by a normaldistribution model with a mean of 5 cm and a standard deviation of 0.02 cm. If a component is chosen atrandoma)What is the probability that the length of this component is between 4.98 and 5.02?

Answer 1

Answer: 0.6826894

Step-by-step explanation:

Given : The length of similar components produced by a company are approximated by a normal distribution model with a $\mu=$ 5 cm and a $\sigma=$0.02 cm.

Let x be the random variable that represents the length of similar components produced by a company.

z-score : $\dfrac{x-\mu}{\sigma}$

For x= 4.98

$\dfrac{4.98-5}{0.02}=-1$

For x= 5.02

$\dfrac{5.02-5}{0.02}=1$

By using the standard z-value table (right -tailed) , the probability that the length of this component is between 4.98 and 5.02 will be :_

$P(4.98$

Hence, the probability that the length of this component is between 4.98 and 5.02= 0.6826894