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The amount of daily time that teenagers spend on a brand A cell phone is normally distributed with a given mean Mu = 2.5 hr and standard deviation Sigma = 0.6 hr. What percentage of the teenagers spend more than 3.1 hr? 5% 10% 16% 32%

Respuesta :

1234alex101

Answer:

Option 3./C. 16%

Step-by-step explanation:

First, we get the z-score by using the equation,

Standard deviation Substituting,

= 1

Z-score = value - mean/

Z-score = (3.1 - 2.5) / 0.6

Converting the z-score to percentage will give us 0.841. Subtracting this value from 1.0 and multiplying The difference by 100%.

Percentage = (1 - 0.0841) x 100%

= 15.9% Thus, 15.9% of the teenagers spend more time In the cellphone.

Didn't you finally want to round it out,

And you get 16%

okpalawalter8

The  percentage of teenagers who spend more than 3.1 hours is mathematically given as

Percentage  = 15.9%

What percentage of teenagers spend more than 3.1 hr?

Question Parameters:

  • Mu = 2.5 hr
  • Sigma = 0.6 hr

Generally, the equation for the  Z-score is mathematically given as

Z-score = value - mean/ standard deviation

Therefore

Z-score = (3.1 - 2.5) / 0.6

Z-score = 0.841

In conclusion

Percentage = (1 - 0.0841) x 100%

Percentage  = 15.9%

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