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Ket [755]
3 years ago
9

based on a survey of 32 randomly selected employees (anonymously, of course) the company has determined that the average amount

of time spent texting over a one-month period is 173 minutes with a standard deviation of 66 minutes. what is the probability that the average amount of time spent using text messages is more than 199 minutes
Mathematics
1 answer:
kramer3 years ago
6 0

Answer: 0.0129

Step-by-step explanation:

Given : Sample size : n=32

The average amount of time spent texting over a one-month period is  : \mu=173\text{ minutes}

Standard deviation : \sigma=66\text{ minutes}

We assume that the time spent texting over a one-month period is normally distributed.

z-score : z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}

For x=  199

z=\dfrac{199-173}{\dfrac{66}{\sqrt{32}}}\approx2.23

Now by using standard normal table, the probability that the average amount of time spent using text messages is more than 199 minutes will be :-

P(x>199)=P(z>2.23)=1-P(z\leq2.23)\\\\=1- 0.9871262=0.0128738\approx0.0129

Hence, the required probability = 0.0129

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The general term for the arithmetic sequence; -13, -11, -9, -7 is Tn = -15 - 2n.

<h3>Arithmetic progression</h3>

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Common difference, d = Second term - first term

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Therefore,

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