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lisabon 2012 [21]

3 years ago

14

Hello! Sorry about My last question! I was lying this is just a Homework assignment and it is completion Credit but if you do it

I will still give you the brainliest!

2 answers:

shutvik [7]3 years ago

4 0

It should be 53 percent or 0.53 :))

Ierofanga [76]3 years ago

3 0

To find the percent of something, turn the percent into a decimal and multiply (move the decimal place 2 places left, e.g. 53% = .53)

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The average time an individual reads online national news reports is 49 minutes. Assume the standard deviation is 16 minutes and

Olin [163]

Answer:

0.118 is the probability someone will spend no more than 30 minutes reading online national news reports.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 49 minutes

Standard Deviation, σ = 16 minutes

We are given that the distribution of time an individual reads is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(no more than 30 minutes)

$P( x \leq 610) = P( z \leq \displaystyle\frac{30 - 49}{16}) = P(z \leq -1.1875)$

Calculation the value from standard normal z table, we have,

$P(x \leq 30) = 0.118 = 11.8\%$

0.118 is the probability someone will spend no more than 30 minutes reading online national news reports.

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