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Suppose that battery life is a normal random variable with μ = 8 and σ = 1.2. using the standard deviation rule, what is the probability that a randomly chosen battery will last between 6.8 and 9.2 hours?

Respuesta :

The probability is 68%. This is how I got it: 

The 68-95-97 rule states that within one standard deviation, 68% of the data is there, within two, 95%, and within 3, 97%. 

The mean is 8, and std. dev. is 1.2. 6.8 is one standard dev to the left, and 9.2 is one standard dev to the right. 

So, the probability that a randomly chosen battery will last between 6.8 and 9.2 hours is 68%

Hope this helps!
Cricetus

The probability that a randomly chosen battery will last between 6.8 as well as 9.2 hrs will be "0.68".

According to the question,

  • μ = 8
  • σ = 1.2

The probability that data values lies within standard deviation i.e.,

→ [tex](\mu -6) (\mu +6)[/tex] will be [tex]0.68[/tex]

The probability that data values lies within two standard deviation i.e.,

→ [tex](\mu -26)(\mu +26)[/tex] will be [tex]0.95[/tex]

The probability that data values lies within three standard deviation i.e.,

→ [tex](\mu -36) (\mu +36)[/tex] will be [tex]0.997[/tex]

Throughout the above examples,

→ [tex]\mu-6 = 8-1.2[/tex]

           [tex]= 6.8[/tex]

→ [tex]\mu +6 = 8+ 1.2[/tex]

            [tex]= 9.2[/tex]

Thus the above answer is correct.

Learn more about probability here:

https://brainly.com/question/13701382

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