Answer:
0.0903
Step-by-step explanation:
Given that :
The mean = 1450
The standard deviation = 220
sample mean = 1560
![P(X > 1560) = P( Z > \dfrac{x - \mu}{\sigma})](https://tex.z-dn.net/?f=P%28X%20%3E%201560%29%20%3D%20P%28%20Z%20%3E%20%5Cdfrac%7Bx%20-%20%5Cmu%7D%7B%5Csigma%7D%29)
![P(X > 1560) = P(Z > \dfrac{1560 - 1450}{220})](https://tex.z-dn.net/?f=P%28X%20%3E%201560%29%20%3D%20P%28Z%20%3E%20%5Cdfrac%7B1560%20-%201450%7D%7B220%7D%29)
![P(X > 1560) = P(Z > \dfrac{110}{220})](https://tex.z-dn.net/?f=P%28X%20%3E%201560%29%20%3D%20P%28Z%20%3E%20%5Cdfrac%7B110%7D%7B220%7D%29)
P(X> 1560) = P(Z > 0.5)
P(X> 1560) = 1 - P(Z < 0.5)
From the z tables;
P(X> 1560) = 1 - 0.6915
P(X> 1560) = 0.3085
Let consider the given number of weeks = 52
Mean
= np = 52 × 0.3085 = 16.042
The standard deviation =
The standard deviation = ![\sqrt {52 \times 0.3085 (1-0.3085)}](https://tex.z-dn.net/?f=%5Csqrt%20%7B52%20%5Ctimes%200.3085%20%281-0.3085%29%7D)
The standard deviation = 3.3306
Let Y be a random variable that proceeds in a binomial distribution, which denotes the number of weeks in a year that exceeds $1560.
Then;
Pr ( Y > 20) = P( z > 20)
![Pr ( Y > 20) = P(Z > \dfrac{20.5 - 16.042}{3.3306})](https://tex.z-dn.net/?f=Pr%20%28%20Y%20%3E%2020%29%20%3D%20P%28Z%20%3E%20%5Cdfrac%7B20.5%20-%2016.042%7D%7B3.3306%7D%29)
![Pr ( Y > 20) = P(Z >1 .338)](https://tex.z-dn.net/?f=Pr%20%28%20Y%20%3E%2020%29%20%3D%20P%28Z%20%3E1%20.338%29)
From z tables
P(Y > 20)
0.0903
Convert the mixed number <span>21 <span>2/3</span></span><span> into an </span><span>improper fraction.
</span>15⋅<span>65/3
</span><span>Cancel 3 and 15.
</span><span>5⋅65
</span>Multiply 5<span> by </span>65<span> to get </span><span>325.
</span>325
Answer:
325
If the E(X) = 1 then that means it has a 100% which is incorrect.
Each
block has an equal probability to contain the error so 1/3 because you
choose 2/6.
A histogram would best represent the data.
Dot plot is a good way to show a certain trend. However, when it comes to the number of a item, in this case the number of students in a certain score range, a histogram will be the most intuitive and concise. Thus, a histogram best represent the data presented here.