Question

The weight of a shaft that is produced by a firm is normally distributed with a mean of 15 pounds. Determine the standard deviation of weight such that 99% of the shafts weighs less than 16.5 pounds. Assume a standard deviation of 0.55 for the shaft. If the firm wants 95% of the shafts it produces to weigh more than 14 pounds, what should be the mean weight for the shafts.

Answer 1

σ = 0.6438

`\mu` = 14.902 pounds

Explanation:

Given that:

The mean = 15

sample mean = 16.5

At 99% C.I, the level of significance is = 1 - 0.99 = 0.01

`Z_(\alpha/2) =Z_(0.01/2)`

`Z_(\alpha/2) =Z_(0.005) =2.33`

By applying the central limit theorem;

`z = (\bar x - \mu)/(\sigma )`

`\sigma = (\bar x - \mu)/(z )`

`\sigma = (16.5 -15)/(2.33)`

σ = 0.6438

Assuming the S.D of the shaft = 0.55

At 95% confidence interval level, from the tables, the z value = -1.64

Using the central limit theorem

`z = (\bar x - \mu)/(\sigma )`

If we make the mean
`\mu` , the subject of the formula: we have:

`\mu =\bar x - z \sigma`

`\mu =14 -(-1.64 * 0.55)`

`\mu =14 -(-0.902)`

`\mu` = 14.902 pounds