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NeTakaya
3 years ago
12

The number of cell phone minutes used by high school seniors follows a normal distribution with a mean of 500 and a standard dev

iation of 50. what is the probability that a student uses more than 580 minutes? a. 0.152 b. 0.0548 c. 0.848 d. 0.903 e. none of these
Mathematics
1 answer:
prohojiy [21]3 years ago
7 0
<span>The number of cell phone minutes used by high school seniors follows a normal distribution with a mean of 500 and a standard deviation of 50. what is the probability that a student uses more than 580 minutes?

Given
&mu;=500
&sigma;=50
X=580
P(x<X)=Z((580-500)/50)=Z(1.6)=0.9452
=>
P(x>X)=1-P(x<X)=1-0.9452=0.0548=5.48%

</span>
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Null hypothesis:\mu \leq 8000  

Alternative hypothesis:\mu > 8000  

z=\frac{8300-8000}{\frac{1200}{\sqrt{64}}}=2  

p_v =P(Z>2)=0.0228  

Step-by-step explanation:

1) Data given and notation  

\bar X=8300 represent the sample mean  

\sigma=1200 represent the population standard deviation  

n=64 sample size  

\mu_o =800 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

2) State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is higher than 8000, the system of hypothesis are :  

Null hypothesis:\mu \leq 8000  

Alternative hypothesis:\mu > 8000  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

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3) Calculate the statistic  

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z=\frac{8300-8000}{\frac{1200}{\sqrt{64}}}=2  

4) P-value  

Since is a one-side upper test the p value would given by:  

p_v =P(Z>2)=0.0228  

5) Conclusion  

If we compare the p value and the significance level assumed, for example \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly higher than $8000.  

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