The mean time it takes for workers at a factory to insert a delicate bolt into an engine is 15 minutes. The standard deviation of time to insert the bolt is 4.0 minutes and the distribution of time is approximately Normally distributed. For a randomly selected worker, what is the approximate probability the bolt will be inserted in 19 minutes or less

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Answer:

required probability is 0.8413

Step-by-step explanation: Given mean [tex]\mu =15[/tex] minutes, standard deviation [tex]\sigma =4[/tex] seconds, we need to find [tex]P\left ( X\leq 19 \right )[/tex]. Using z score

[tex]z=\frac{X-\mu }{\sigma }\\=\frac{19-15}{4}\\=1[/tex]

So,

[tex]P\left ( X\leq 19 \right )\\=P\left ( Z\leq 1 \right )\\=0.8413[/tex]

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