Question

Can you answer the following showing formula used and all workings out. The length of a...

Can you answer the following showing formula used and all workings out.

The length of a screw manufactured by a company is normally distributed with mean 2.5cm and a standard deviation 0.1cm. Specifications call for the lengths to range from 2.4cm to 2.6cm.

1, what value of standard deviation is required so that less than 7% of screws have a length greater than 2.8cm?

2, another manufacturer produces the same screw for which 15% of the screws have length less than 1.75cm and 12% of the screws have length higher than 3.2cm. Find the mean and standard deviation of the length of the screw manufactured by the company assuming the length is normally distributed.

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Answer #1

1.

The mean length of the screw is 2.5 cm.

Let be the value of standard deviation is required so that less than 7% of screws have a length greater than 2.8cm.

P(X > 2.8) = 0.07

P[Z > (2.8 - 2.5)/] = 0.07

0.3/ = 1.476 (Using Standard normal distribution tables)

= 0.3 / 1.476 = 0.2033 cm.

2.

Let and be the length of the screw manufactured by the company.

Probability that screws have length less than 1.75cm = 0.15

P(X < 1.75) = 0.15

P[Z < (1.75 - ) / ] = 0.15

(1.75 - ) / = -1.036 ---(1)     (Using Standard normal distribution tables)

Probability that screws have length higher than 3.2 cm = 0.12

P(X > 3.2) = 0.12

P[Z < (3.2 - ) / ] = 1.175

(3.2 - ) / = 1.175 ---(2)    (Using Standard normal distribution tables)

Dividing (1) by (2), we get

(1.75 - ) / (3.2 - ) = -1.036 / 1.175

1.75 - = -0.8817 * (3.2 - )

1.75 - = -2.82144 + 0.8817

1.8817 = 1.75 + 2.82144

= 4.57144 / 1.8817 = 2.4294 cm

From (1),

(1.75 - 2.4294) / = -1.036

= -0.6794 / -1.036 = 0.6558 cm

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