Answer:
![P(X](https://tex.z-dn.net/?f=%20P%28X%20%3C9%29%20)
And we can use the cumulative distribution function given by:
![F(X) = \frac{x-a}{b-a}, a\leq X \leq b](https://tex.z-dn.net/?f=%20F%28X%29%20%3D%20%5Cfrac%7Bx-a%7D%7Bb-a%7D%2C%20a%5Cleq%20X%20%5Cleq%20b)
And using this we got:
![P(X](https://tex.z-dn.net/?f=%20P%28X%20%3C9%29%20%3D%20F%289%29%20%3D%20%5Cfrac%7B9-3%7D%7B11-3%7D%3D%200.75)
Step-by-step explanation:
For this case we assume that X= represent the waiting times and for this case we have the following distribution:
![X \sim Unif (a= 3, b =11)](https://tex.z-dn.net/?f=%20X%20%5Csim%20Unif%20%28a%3D%203%2C%20b%20%3D11%29)
And the expected value is given by:
![\mu= E(X) = \frac{a+b}{2}= \frac{3+11}{2}=7](https://tex.z-dn.net/?f=%5Cmu%3D%20E%28X%29%20%3D%20%5Cfrac%7Ba%2Bb%7D%7B2%7D%3D%20%5Cfrac%7B3%2B11%7D%7B2%7D%3D7)
And the variance is given by:
![Var(X) \sigma^2 = \frac{(b-a)^2}{12} = \frac{(11-3)^2}{12} = 5.333](https://tex.z-dn.net/?f=%20Var%28X%29%20%5Csigma%5E2%20%3D%20%5Cfrac%7B%28b-a%29%5E2%7D%7B12%7D%20%3D%20%5Cfrac%7B%2811-3%29%5E2%7D%7B12%7D%20%3D%205.333)
And we can find the deviation like this:
![Sd(X) = \sqrt{5.333}= 2.309](https://tex.z-dn.net/?f=%20Sd%28X%29%20%3D%20%5Csqrt%7B5.333%7D%3D%202.309)
And we want to find this probability:
![P(X](https://tex.z-dn.net/?f=%20P%28X%20%3C9%29%20)
And we can use the cumulative distribution function given by:
![F(X) = \frac{x-a}{b-a}, a\leq X \leq b](https://tex.z-dn.net/?f=%20F%28X%29%20%3D%20%5Cfrac%7Bx-a%7D%7Bb-a%7D%2C%20a%5Cleq%20X%20%5Cleq%20b)
And using this we got:
![P(X](https://tex.z-dn.net/?f=%20P%28X%20%3C9%29%20%3D%20F%289%29%20%3D%20%5Cfrac%7B9-3%7D%7B11-3%7D%3D%200.75)
Answer:
points needed hope you under stand
Step-by-step explanation:
stay safe thoughω
Work hours per day =
![\frac{21}{4} = 5 \frac{1}{4}](https://tex.z-dn.net/?f=%20%5Cfrac%7B21%7D%7B4%7D%20%20%3D%205%20%5Cfrac%7B1%7D%7B4%7D%20)
Therefore, Sara works for 5 hours and 15 minutes per day
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