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svet-max [94.6K]
3 years ago
8

Steel rods are manufactured with a mean length of 29 centimeter (cm). Because of variability in the manufacturing process, the l

engths of the rods are approximately normally distributed with a standard deviation of 0.07 cm. (a) What proportion of rods has a length less than 28.9 cm? (b) b) Any rods that are shorter than 24.84 cm or longer than 25.16 cm are discarded. What proportion of rods will be discarded?
Mathematics
1 answer:
zavuch27 [327]3 years ago
4 0

Solution :

Given data :

The mean length of the steel rod = 29 centimeter (cm)

The standard deviation of a normally distributed lengths of rods = 0.07 centimeter (cm)

a). We are required to find the proportion of rod that have a length of less than 28.9 centimeter (cm).

Therefore, P(x < 28.9) = P(z < (28.9-29) / 0.07)

                                    = P(z < -1.42)

                                   = 0.0778

b). Any rods which is shorter than 24.84 cm or longer than 25.16 cm that re discarded.

Therefore,

P (x < 24.84 or 25.16 < x) = P( -59.42 < z or -54.85)

                                         = 1.052

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Step-by-step explanation:

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P(X < 8.52) = 64.8%

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A multiplicative inverse of an integer a is an integer x such that the product ax is congruent to 1 with respect to the modulus m.  

1. Z_{11}=\{\overline{0},\overline{1},\overline{2},\overline{3},\overline{4},\overline{5},\overline{6},\overline{7},\overline{8},\overline{9},\overline{10}\}.

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2.   Z_{12}=\{\overline{0},\overline{1},\overline{2},\overline{3},\overline{4},\overline{5},\overline{6},\overline{7},\overline{8},\overline{9},\overline{10},\overline{11}\}.

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The multiplicative inverse of 5 in Z_{12} is 5.

3.  Z_{13}=\{\overline{0},\overline{1},\overline{2},\overline{3},\overline{4},\overline{5},\overline{6},\overline{7},\overline{8},\overline{9},\overline{10},\overline{11},\overline{12}\}.

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  • 5\cdot 9=45=\overline{6};
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