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The mean playing time for a large collection of compact discs is 34 minutes, and the standard deviation is 6 minutes. (a) What value (in minutes) is 1 standard deviation above the mean

Respuesta :

Chrisnando

Answer:

40 minutes

Step-by-step explanation:

Let's take the average playing time as u, and let the standard deviation be [tex] \sigma [/tex].

Thus,

[tex] u = 34 [/tex]

[tex] \sigma = 6 [/tex]

To find the value of 1 standard deviation above the mean, we have:

[tex] u + 1\sigma [/tex]

Since u = 34 minutes and [tex]1\sigma [/tex] = 6 minutes, we now have:

[tex] = 34 + 6 = 40 [/tex]

The value (in minutes) of 1 standard deviation above the mean is 40 minutes.

ajeigbeibraheem

Answer:

40 minutes

Step-by-step explanation:

Given that:

The average time playing time of compact disc in a large collection= 34 minutes and the standard deviation = 6 minutes.

Let [tex]\mu[/tex] represent the average playing time and [tex]\sigma[/tex] to be the standard deviation.

So;

[tex]\mu = 34 \\ \\ \sigma = 6[/tex]

The value for 1 standard deviation above the mean is;

[tex]x+1 s= 34+6 \\ \\ = 40[/tex]

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