

# we have
P_mean = 100
P_std = 15
S_mean = 108
Sample_no = 36

H<sub>0<sub> : mu = 100

H<sub>1<sub> : mu != 100

let say , alpha = 0.05

\begin{align}
Z = (X - mu) / (std / \sqrt{n}
\end{align}
\begin{align}
Z_{0.05} = 1.96
\end{align}
\begin{align}
Z = (108 - 100) / (15 / \sqrt{36} = 3.2
\end{align}


since Z > Z_{0.05} i.e. 3.2 > 1.96 , we will reject the null hypothesis at 5% level of significance. 




let P1 be the probability for Repulicans first state and P2 be the Probabiity for Repulicans for second state.
hence, P1 = 52% = 0.52
       P2 = 47% = 0.47

sample size = 100

so the difference in proportion between two state be p

P = P1 - P2
  = 0.52 - 0.47
  = 0.05

calculating the variance for Republicans.
                  
P<sub>variance<sub>   = P1(1-P1)/sample_size + P2(1-P2)/sample_size
      = 0.52(1-0.52)/100 + 0.47(1-0.47)/100
      = (0.52 * 0.48 + 0.47 * 0.53)/100   
      = 0.00498

std = sqrt(0.004987) = 0.071
the difference between the first and second state of Republicans need to less than 0 in orer to achieve the greater percentage in second state. 
hence the the valud of the difference is aproxmetly equal to 0 so p ~ 0 

hence,
 
 Z = 0 - 0.05/ 0.071 = -0.704
 using Z score table , the value for -704 = 0.239
 thus the probability for greater percentage of Republicans in second state than in first state is 23.9 %





We have ,

X = 1100
mu = 1026
std = 209 
hence calculating the Z-score value 

Z = (1100-1026)/209 = 0.354

hence, te score is 35.4 5 standard deviation above the mean score in SAT. 

using the z-score table for z(0.354)  = 0.636. ie. 63.6% chnage that the score is more than the average test takers.


```python

```

7323dd721da27e2b7d7ef7790d913cbe3d6c6ae6
ipynb
Jupyter Notebook
stats3.ipynb
Rankush888/IneuronMLDLassignment
ba3184b1f19de27dcdf8bb7cc040fa71bbbe7dbe
MIT
stats3.ipynb
Rankush888/IneuronMLDLassignment
ba3184b1f19de27dcdf8bb7cc040fa71bbbe7dbe
MIT
stats3.ipynb
Rankush888/IneuronMLDLassignment
ba3184b1f19de27dcdf8bb7cc040fa71bbbe7dbe
MIT
Qwen/Qwen-72B
1. YES
2. YES