Respuesta :
Answer:
a) [tex]P(W=62)=(90C62)(0.5)^{62} (1-0.5)^{90-62}=0.000125[/tex]
b) [tex]P(W\leq 62)=P(\frac{W-\mu}{\sigma}\leq \frac{62-45}{4.743})=P(Z\leq 3.584)=0.9998[/tex]
c) [tex]P(50 \leq W \leq 60)=P(Z<3.163)-P(Z<1.054)=0.9992-0.854=0.1451[/tex]
Step-by-step explanation:
1) Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Solution to the problem
Let W the random variable of interest, on this case we now that:
[tex]W \sim Binom(n=90, p=0.5)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(W)=(nCw)(p)^w (1-p)^{n-w}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCw=\frac{n!}{(n-w)! w!}[/tex]
Part a
For this case we want this probability:
[tex]P(W=62)=(90C62)(0.5)^{62} (1-0.5)^{90-62}=0.000125[/tex]
Part b
We need to check the conditions in order to use the normal approximation.
[tex]np=90*0.5=45 \geq 10[/tex]
[tex]n(1-p)=90*(1-0.5)=45 \geq 10[/tex]
So we see that we satisfy the conditions and then we can apply the approximation.
If we appply the approximation the new mean and standard deviation are:
[tex]E(W)=np=90*0.5=45[/tex]
[tex]\sigma=\sqrt{np(1-p)}=\sqrt{90*0.5(1-0.5)}=4.743[/tex]
And we can use the z score given by this formula:
[tex]z=\frac{w-\mu}{\sigma}[/tex]
[tex]P(W\leq 62)[/tex]
[tex]P(W\leq 62)=P(\frac{W-\mu}{\sigma}\leq \frac{62-45}{4.743})=P(Z\leq 3.584)=0.9998[/tex]
Part c
For this case we want this probability:
[tex]P(50 \leq W \leq 60)=P(\frac{50-45}{4.743} \leq \frac{W-\mu}{\sigma} \leq \frac{60-45}{4.743}[/tex]
[tex]P(1.054 \leq Z \leq 3.163)=P(Z<3.163)-P(Z<1.054)=0.9992-0.854=0.1451[/tex]
Answer:
a) 0.000125
b) 0.9998
c) 0.1706
Step-by-step explanation:
W has binomial distribution with n = 90, and p = 0.5.
The mean of W is 0.5*100 = 45, and the variance is 0.5*(1-0.5)*90 = 22.5, so the standard deviation is √22.5.
As a consecuence of the Central Limit Theorem, W has approximately Normal distribution X ≈ N(μ=45,σ=√22.5).
To solve a, we can calculate the exact probability using the binomial formula.
[tex] P(W=62) = {90 \choose 62} * (0.5)^62 * (1-0.5)^28 = 0.000125 [/tex]
b) We have to make a correction for continuity (because P(X=62) = 0). We standarize X to obtain a random variable Y with distribution N(0,1)
P(W≤62) = P( W < 62.5) = P(X < 62.5) = P( (X -45)/√2.5 < 62.5-45/√22.5) = P(Y < 3.689)
Where Y = (X -45)/√22.5. Y has distribution N(0,1) and the values of the cummulative distribution function of Y, Φ, are tabulated. You can find the values of Φ on the attached file. P(Y < 3.689) = Φ(3.689) = 0.9998. This shows that the number of successful attemps is unlikely to be far from its mean 45.
c) P(50 < W < 60) = P(49.5 < W < 60.5) (correction of continuity) = P(49.5 < X < 60.5) = P((49.5-45)/√22.5 < Y < (60.5-45)/√22.5) = P(0.94868 < Y < 3.26768) = Ф(3.2678)-Ф(0.94868) = 0.9995-0.8289 = 0.1706
