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Answer with explanation</u>
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Let 
be the average life of light bulbs.
As per given , we have
Null hypothesis : 
Alternative hypothesis : 
Since 
is right-tailed and population standard deviation is also known, so we perform right-tailed z-test.
Formula for Test statistic : 
where, n= sample size
= sample mean
= Population mean
=population standard deviation
For 
, we have
Using z-value table , Critical one-tailed test value for 0.06 significance level :
Decision : Since critical z value (1.5548) < test statistic (1.6180), so we reject the null hypothesis .
[We reject the null hypothesis when critical value is less than the test statistic value .]
Conclusion : We have enough evidence at 0.06 significance level to support the claim that the new filament yields a longer bulb life
Answer:
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Answer:
a) P(x<5)=0.
b) E(X)=15.
c) P(8<x<13)=0.3.
d) P=0.216.
e) P=1.
Step-by-step explanation:
We have the function:
a) We calculate the probability that you need less than 5 minutes to get up:
Therefore, the probability is P(x<5)=0.
b) It takes us between 10 and 20 minutes to get up. The expected value is to get up in 15 minutes.
E(X)=15.
c) We calculate the probability that you will need between 8 and 13 minutes:
![P(8\leq x\leq 13)=P(10\leqx\leq 13)\\\\P(8\leq x\leq 13)=\int_{10}^{13} f(x)\, dx\\\\P(8\leq x\leq 13)=\int_{10}^{13} \frac{1}{10} \, dx\\\\P(8\leq x\leq 13)=\frac{1}{10} \cdot [x]_{10}^{13}\\\\P(8\leq x\leq 13)=\frac{1}{10} \cdot (13-10)\\\\P(8\leq x\leq 13)=\frac{3}{10}\\\\P(8\leq x\leq 13)=0.3](https://tex.z-dn.net/?f=P%288%5Cleq%20x%5Cleq%2013%29%3DP%2810%5Cleqx%5Cleq%2013%29%5C%5C%5C%5CP%288%5Cleq%20x%5Cleq%2013%29%3D%5Cint_%7B10%7D%5E%7B13%7D%20f%28x%29%5C%2C%20dx%5C%5C%5C%5CP%288%5Cleq%20x%5Cleq%2013%29%3D%5Cint_%7B10%7D%5E%7B13%7D%20%5Cfrac%7B1%7D%7B10%7D%20%5C%2C%20dx%5C%5C%5C%5CP%288%5Cleq%20x%5Cleq%2013%29%3D%5Cfrac%7B1%7D%7B10%7D%20%5Ccdot%20%5Bx%5D_%7B10%7D%5E%7B13%7D%5C%5C%5C%5CP%288%5Cleq%20x%5Cleq%2013%29%3D%5Cfrac%7B1%7D%7B10%7D%20%5Ccdot%20%2813-10%29%5C%5C%5C%5CP%288%5Cleq%20x%5Cleq%2013%29%3D%5Cfrac%7B3%7D%7B10%7D%5C%5C%5C%5CP%288%5Cleq%20x%5Cleq%2013%29%3D0.3)
Therefore, the probability is P(8<x<13)=0.3.
d) We calculate the probability that you will be late to each of the 9:30am classes next week:
![P(x>14)=\int_{14}^{20} f(x)\, dx\\\\P(x>14)=\int_{14}^{20} \frac{1}{10} \, dx\\\\P(x>14)=\frac{1}{10} [x]_{14}^{20}\\\\P(x>14)=\frac{6}{10}\\\\P(x>14)=0.6](https://tex.z-dn.net/?f=P%28x%3E14%29%3D%5Cint_%7B14%7D%5E%7B20%7D%20f%28x%29%5C%2C%20dx%5C%5C%5C%5CP%28x%3E14%29%3D%5Cint_%7B14%7D%5E%7B20%7D%20%5Cfrac%7B1%7D%7B10%7D%20%5C%2C%20dx%5C%5C%5C%5CP%28x%3E14%29%3D%5Cfrac%7B1%7D%7B10%7D%20%5Bx%5D_%7B14%7D%5E%7B20%7D%5C%5C%5C%5CP%28x%3E14%29%3D%5Cfrac%7B6%7D%7B10%7D%5C%5C%5C%5CP%28x%3E14%29%3D0.6)
You have 9:30am classes three times a week. So, we get:
Therefore, the probability is P=0.216.
e) We calculate the probability that you are late to at least one 9am class next week:
![P(x>9.5)=\int_{10}^{20} f(x)\, dx\\\\P(x>9.5)=\int_{10}^{20} \frac{1}{10} \, dx\\\\P(x>9.5)=\frac{1}{10} [x]_{10}^{20}\\\\P(x>9.5)=1](https://tex.z-dn.net/?f=P%28x%3E9.5%29%3D%5Cint_%7B10%7D%5E%7B20%7D%20f%28x%29%5C%2C%20dx%5C%5C%5C%5CP%28x%3E9.5%29%3D%5Cint_%7B10%7D%5E%7B20%7D%20%5Cfrac%7B1%7D%7B10%7D%20%5C%2C%20dx%5C%5C%5C%5CP%28x%3E9.5%29%3D%5Cfrac%7B1%7D%7B10%7D%20%5Bx%5D_%7B10%7D%5E%7B20%7D%5C%5C%5C%5CP%28x%3E9.5%29%3D1)
Therefore, the probability is P=1.
Answer:
-3x^2-22x-7
Step-by-step explanation:
