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qwelly [4]
3 years ago
8

\The lengths of a lawn mower part are approximately normally distributed with a given mean Mu = 4 in. and standard deviation Sig

ma = 0.2 in. What percentage of the parts will have lengths between 3.8 in. and 4.2 in.?
Mathematics
1 answer:
chubhunter [2.5K]3 years ago
3 0

Answer:

68.26% of the parts will have lengths between 3.8 in. and 4.2 in.

Step-by-step explanation:

The lengths of a lawn mower part are approximately normally distributed

\mu = 4

Standard deviation =\sigma =0.2

We are supposed to find  What percentage of the parts will have lengths between 3.8 in. and 4.2 in . i.e. P(3.8<x<4.2)

Formula : Z=\frac{x-\mu}{\sigma}

At x = 3.8

Z=\frac{3.8-4}{0.2}

Z=-1

Refer the z table for p value

p value =0.1587

At x =4.2

Z=\frac{4.2-4}{0.2}

Z=1

Refer the z table for p value

p value =0.8413

P(3.8<x<4.2)=P(x<4.2)-P(x<3.8)=0.8413-0.1587=0.6826

So, 68.26% of the parts will have lengths between 3.8 in. and 4.2 in.

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