Respuesta :
Answer:
Explanation:
given,
Mean,μ= 35mm
Standard Deviation,σ = 0.5mm
Sample size, n = 36
Sample Standard deviation =[tex]\dfrac{\sigma}{\sqrt{n}}[/tex]
= [tex]\dfrac{0.5}{6}[/tex]
= 0.0833
The interested diameter is between 34.95 to 35.18 mm
Calculating the Z score of the for the diameter mentioned.
[tex]P(\dfrac{x_1-\mu}{\sigma})<Z<\dfrac{x_2-\mu}{\sigma}[/tex]
[tex]P(\dfrac{34.95-35}{0.0833})<Z<\dfrac{35.18-35}{0.0833}[/tex]
[tex]-0.6< Z < 2.16[/tex]
now, Form Z-table
[tex] P(Z<-0.6) = 0.2741[/tex]
[tex]P(Z<2.16) = 0.9846[/tex]
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.
