Question

A tire manufacturer is measuring the margin of error in the thickness of their tires to make sure it is within safety limits. Overall, the tires' thickness is normally distributed with a mean of 0.45 inches and a standard deviation of 0.05 inches. What thickness separates the lowest 5% of the means from the highest 95% in a sample size of 65 tires

Answer 1

Answer:

Z = -1.65

$\bar x \approx 0.44 \ inches$

Step-by-step explanation:

The main objective is to compute the data for the Z value and determine the $\bar x$ of the sample distribution

Given that;

the tires' thickness is normally distributed  with a mean μ = 0.45 in

standard deviation σ = 0.05 in

sample size  =  65 tires

Also; we are being told that the  thickness separates the lowest 5% of the means from the highest 95%

∴

P(Z < Z) =0.05

From the Z- table

P(Z < -1.645) = 0.05

Z = -1.65

Similarly;

Let consider $\bar x$  to be the sample mean;

Then:

mean $(\mu_{\bar x}) = \mu = 0.45$

standard deviation$(\sigma_{\bar x} ) = \dfrac{\sigma}{\sqrt{n}}$

$=\dfrac{0.05}{\sqrt{65}}$

= 0.00620174

By applying the Z-score formula:

x =  μ + ( Z × σ )

$\bar x = \mu _{\bar x} +(Z * \sigma _{\bar x})$

$\bar x = 0.45 + (-1.65 *0.00620174)$

$\bar x= 0.439767129$

$\bar x \approx 0.44 \ inches$