Question

The lengths of a lawn mower part are approximately normally distributed with a given mean Mu = 4 in. and standard deviation Sigma = 0.2 in. What percentage of the parts will have lengths between 3.8 in. and 4.2 in.?


68%

95%

99.7%

34%

Answer 1

68% of the parts will have lengths between 3.8 inches and 4.2 inches.

Step-by-step explanation:

We are given that the lengths of a lawn mower part are approximately normally distributed with a given mean Mu = 4 inches and standard deviation Sigma = 0.2 inches.

Let X = lengths of a lawn mower part

SO, X ~ Normal($\mu=4,\sigma^{2} =0.2^{2}$)

The z-score probability distribution of normal distribution is given by;

                Z =  $\frac{X-\mu}{\sigma}$  ~ N(0,1)

where, $\mu$ = population mean = 4 inches

            $\sigma$ = standard deviation = 0.2 inches

So, the percentage of the parts that will have lengths between 3.8 and 4.2 is given by = P(3.8 inches < X < 4.2 inches)

 P(3.8 < X < 4.2) = P(X < 4.2) - P(X $\leq$ 3.8)

 P(X < 4.2) = P( $\frac{X-\mu}{\sigma}$ < $\frac{4.2-4}{0.2}$ ) = P(Z < 1) = 0.84134

 P(X $\leq$ 12.85 mm) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{3.8-4}{0.2}$ ) = P(Z $\leq$ -1) = 1 - P(Z < 1)

                                                         = 1 - 0.84134 = 0.15866

The above probabilities are calculated using z table by looking at the critical values of x = 1 which gives an probability area of 0.84134.

Therefore, P(3.8 < X < 4.2) = 0.84134 - 0.15866 = 0.6827 or 68.3%

Hence, 68% of the parts will have lengths between 3.8 inches and 4.2 inches.

Learn more about normal distribution questions;

brainly.in/question/15727042

Answer 2

Answer:

It's B - 68%

Step-by-step explanation:

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