Question

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.

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