Answer :
Answer: 0.6827
Step-by-step explanation:
Let x be the random variable that represents the length of similar components produced by a company.
Given : The length of similar components produced by a company follows a normal distribution with [tex]\mu=5\ cm[/tex] and [tex]\sigma=0.02\ cm[/tex].
Using formula : [tex]z=\dfrac{x-\mu}{\sigma}[/tex]
Z-score corresponds to x= 4.98
[tex]z=\dfrac{4.98-5}{0.02}=-1[/tex]
Z-score corresponds to x= 5.02
[tex]z=\dfrac{5.02-5}{0.02}=1[/tex]
The probability that the length of this component is between 4.98 and 5.02 cm.:
[tex]P(4.98<x<5.02)=P(-1<z<1)\\\\=1-2P(z>|1|)\\\\=1-2(0.1586553)[/tex] [using z-table for right tailed test]
[tex]=0.6826894\approx0.6827[/tex]
Hence, the probability that the length of this component is between 4.98 and 5.02 cm 0.6827=0.6827