Question

The length of similar components produced by a company follows a normal distribution with a mean of 5 cm and a standard deviation of 0.02 cm. If a component is chosen at random, what is the probability that the length of this component is between 4.98 and 5.02cm..?

Answer 1

Answer: 0.6827

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
`\mu=5\ cm` and
`\sigma=0.02\ cm`.

Using formula :
`z=(x-\mu)/(\sigma)`

Z-score corresponds to x= 4.98

`z=(4.98-5)/(0.02)=-1`

Z-score corresponds to x= 5.02

`z=(5.02-5)/(0.02)=1`

The probability that the length of this component is between 4.98 and 5.02 cm.:

`P(4.98<x<5.02)=P(-1<z<1)\\\\=1-2P(z>|1|)\\\\=1-2(0.1586553)` [using z-table for right tailed test]

`=0.6826894\approx0.6827`

Hence, the probability that the length of this component is between 4.98 and 5.02 cm 0.6827=0.6827