Answer: 0.6826894
Explanation:
Given : The length of similar components produced by a company are approximated by a normal distribution model with a
5 cm and a
0.02 cm.
Let x be the random variable that represents the length of similar components produced by a company.
z-score :

For x= 4.98

For x= 5.02

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<x<62000)=P(-1<z<1)=1-2P(z>|1|)\ \ \\\\=1-2P(Z>|z|)]\\\\=1-2(0.1586553)\\\\=0.6826894](https://img.qammunity.org/2020/formulas/mathematics/high-school/gga55epxe0r5junvw1plpijtwe8kcijo6t.png)
Hence, the probability that the length of this component is between 4.98 and 5.02= 0.6826894