====== Wnioskowanie Statystyczne - Lista 7. ======
===== Zadanie 1. =====
$n = 10$\\
$\mu = 0.25$\\
$\sigma^2 = 0.1875$\\
\\
$(0.25)^6 \cdot (0.75)^4 = 7.724761962890625 * 10^{-5}$\\
\\
$P( X \leq 7.5 ) = P( \frac {X - n\mu} { \sqrt {n \cdot \sigma^2}} \leq \frac {7.5 - 2.5} { \sqrt{ 10*0.1875 } } ) = \Phi( \frac 5 {1.3693063937629153} ) = \Phi( 3.6514837167011076 ) =0.9998696351835723$\\

Dyskretnie: 0.99958419799804688
<code python>
def probdisc( dens, args ):
    return reduce( (lambda a, b: a+b), map( dens, args) )

z1 = (lambda x:scipy.stats.binom.pmf(x,10,0.25))

probdisc( ws.z1, range(0,8) )
</code>

\\
$P( X \leq k ) = P( \frac {X - n\mu} { \sqrt {n \cdot \sigma^2}} \leq \frac {k - 2.5} { \sqrt{ 10*0.1875 } } ) = \Phi( \frac {k-2.5} {1.3693063937629153} ) \geq 0.7$\\
Z tabeli:
$\Phi(0.53) \sim 0.7019$\\
$\frac {k-2.5} {1.3693063937629153} \geq 0.53$\\
$k \geq 0.53 \cdot 1.3693063937629153 + 2.5$\\
$k \geq 3.225732388694345$\\

===== Zadanie 2. =====
===== Zadanie 3. =====
===== Zadanie 4. =====
$n = 1000$\\
$\mu = 0.8$\\
$\sigma^2 = 0.16$\\
$P( X > 800 ) = 1 - P( X \leq 800 ) = 1 - P( \frac {X-n\cdot\mu} {\sqrt{n\cdot\sigma^2}} \leq \frac {800-800} {40} ) = 1 - \Phi(0.5) = \frac 1 2$\\
$P( X < 900 ) = P( X \leq 899.5 ) =P( \frac {X-n\cdot\mu} {\sqrt{n\cdot\sigma^2}} \leq \frac {899.5-800} {40} ) =\Phi(2.4875) =0.9934$\\
$P( 700 \leq X \leq 800 ) = P( X \leq 800.5 ) - P( X < 700 ) = P( Z \leq \frac {0.5} {40} ) - P( Z \leq -\frac {100.5}{40} )  =
\Phi(\frac 1 {80}) - \Phi(-2.5125) = 0.5 - 0.0060 = 0.49399999999999999$\\
<code python>
z4 = (lambda x:scipy.stats.binom.pmf(x,1000,0.8))

z41 = probdisc( z4, range(800,1001) )
z42 = probdisc( z4, range(0,900) )
z43 = probdisc( z4, range(700,801) )

>>> ws.z41
0.5189114355580372
>>> ws.z42
0.99999999999999989
>>> ws.z43
0.51261392561400121
</code>

===== Zadanie 5. =====
$n = 100$\\
$\mu = 50$\\
$\sigma^2 = 16$\\
$P( X > 4500 ) = 
1 - P( X \leq 4500 ) =
1 - P( Z \leq \frac {4500 - 5000} {40} ) = 1 - \Phi( -12.5 ) = 1 - 3.73256 \cdot 10^{-36} \sim 1$\\
===== Zadanie 6. =====
$n = 100$\\
$\mu = 3.5$\\
$\sigma^2 = \frac {35} {12}$\\
$P( X > 390 ) = 1 - P( X \leq 390 ) = 1 - \Phi( \frac {390 - 350} {\sqrt{\frac{3500}{12}}}) = 1 - \Phi( \frac {40} {17.07825127659933} ) = 1 - \Phi( 2.3421601750764793 )=0.0095862423776115202$\\
$P( X \leq 399.5 ) = \Phi( \frac {399.5 - 350} {\sqrt{\frac{3500}{12}}})= \Phi(2.8984232166571435)=0.99812477935759081
$\\
$P( 320 \leq X \leq 390 ) = P( X \leq 390) - P( X \leq 319.5) = \Phi( \frac {390 - 350} {\sqrt{\frac{3500}{12}}}) - \Phi( \frac {319.5 - 350} {\sqrt{\frac{3500}{12}}}) =  \Phi( 2.3421601750764793 ) - \Phi(-1.7858971334958156)=0.95335580002063425$\\
===== Zadanie 7. =====
$n = 100$\\
$\mu = 0.9$\\
$\sigma^2 = 0.09$\\
$P( X > 85 ) = 1 - P( X \leq 85 )=\Phi( \frac {-5} {3})=0.022750131948179195$\\
===== Zadanie 8. =====
<code python>
import math

p = 0.15865525393145707
q = 1 - p

sil = math.factorial

def bico(a,b):
    return sil(a)/(sil(b)*sil(a-b))

def prob(a,b):
    return bico(a,b)*(p**a)*(q**b)


</code>

$\mu = 10$\\
$\sigma^2 = 1$\\

$(a)$\\
$p = P( X \leq 9 ) = \Phi( -1 )=0.15865525393145707$

$(b) $\\
$P = {5 \choose 2 } \cdot p^{2}\cdot q^{3}=0.0007115743202948143$\\
$(c ) $\\
$P = {50 \choose 20 } \cdot p^{20}\cdot q^{30}=1.5685300628522175e-28 $\\
$(d) $\\
$P = {500 \choose 200 } \cdot p^{200}\cdot q^{300}= 0$\\
===== Zadanie 9. =====
$n = 100$\\
$\mu = 0.1$\\
$\sigma^2 = 0.09$\\
$P(9.8\leq X \leq 10.2) = P(X\leq 10.2) - P(X \leq 9.8)= \Phi(\frac{0.2}{\sqrt{90}}) - \Phi(\frac{-0.2}{\sqrt{90}})=0.016819637571827251$

<code python>
import scipy.stats
cdf = scipy.stats.norm.cdf
cdf( 0.2/(math.sqrt(90))) - cdf((-0.2)/(math.sqrt(90)))
</code>
===== Zadanie 10. =====
$\mu = 0.36$\\
$P(0.24\leq X \leq 0.48)\leq 0.1 $\\
$P( X \leq 0.48) - P( X \leq 0.24) \leq 0.1 $\\
$\displaystyle \Phi(\frac {0.48-n*0.36}{\sqrt{n*n*0.36*0.64}}) - \Phi(\frac {0.24-n*0.36}{\sqrt{n*n*0.36*0.64}}) \leq 0.1$

<code python>
import math
import scipy.stats

cdf = scipy.stats.norm.cdf

def guess(n):
    return cdf( (0.48-n*0.36)/math.sqrt(n*n*0.36*0.64)) - cdf( (0.24-n*0.36)/math.sqrt(n*n*0.36*0.64))
</code>

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