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?

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

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

z-score : [tex]\dfrac{x-\mu}{\sigma}[/tex]

For x= 4.98

[tex]\dfrac{4.98-5}{0.02}=-1[/tex]

For x= 5.02

[tex]\dfrac{5.02-5}{0.02}=1[/tex]

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 :_

[tex]P(4.98<x<62000)=P(-1<z<1)=1-2P(z>|1|)\ \ \\\\=1-2P(Z>|z|)]\\\\=1-2(0.1586553)\\\\=0.6826894[/tex]

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