Question

The diameter of bearingsis known to have a mean of 35 mm with a standard deviation of 0.5 mm. A random sample of 36 bearings is selected. What is the probability that the average diameter of these selected bearings will be between 34.95 and 35.18 mm?

Answer 1

Answer:

Explanation:

given,

Mean,μ= 35mm  

Standard Deviation,σ = 0.5mm  

Sample size, n = 36  

Sample Standard deviation =$\dfrac{\sigma}{\sqrt{n}}$

                                             = $\dfrac{0.5}{6}$

                                             = 0.0833

The interested diameter is between 34.95 to 35.18 mm  

Calculating the Z score of the for the diameter mentioned.

$P(\dfrac{x_1-\mu}{\sigma})$

$P(\dfrac{34.95-35}{0.0833})$

$-0.6< Z < 2.16$

now, Form Z-table

$P(Z$

$P(Z$

Subtracting the value

      = 0.9846 - 0.2741

      = 0.71

Hence, the required probability is that the diameter of bearing is in between  34.95 and 35.18 mm is equal to 0.71.