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rosijanka [135]
3 years ago
9

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?
Business
1 answer:
timama [110]3 years ago
6 0

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.

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