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

Answer:

σ = 0.6438

$\mu$ = 14.902  pounds

Step-by-step 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;

$z = \dfrac{\bar x - \mu}{\sigma }$

$\sigma = \dfrac{\bar x - \mu}{z }$

$\sigma = \dfrac{16.5 -15}{2.33}$

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

$z = \dfrac{\bar x - \mu}{\sigma }$

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

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

$\mu =14 -(-1.64 \times 0.55)$

$\mu =14 -(-0.902)$

$\mu$ = 14.902  pounds