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Sagot :
Using the distributions, it is found that there is a:
a) 0.1762 = 17.62% probability that 10 of the tosses will fall heads and 10 will fall tails.
b) 0% probability that 10 of the tosses will fall heads and 10 will fall tails.
c) 0.1742 = 17.42% probability that 10 of the tosses will fall heads and 10 will fall tails.
Item a:
Binomial probability distribution
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- 20 tosses, hence [tex]n = 20[/tex].
- Fair coin, hence [tex]p = 0.5[/tex].
The probability is P(X = 10), thus:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 10) = C_{20,10}.(0.5)^{10}.(0.5)^{10} = 0.1762[/tex]
0.1762 = 17.62% probability that 10 of the tosses will fall heads and 10 will fall tails.
Item b:
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
- The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with [tex]\mu = np, \sigma = \sqrt{np(1-p)}[/tex].
The probability of an exact value is 0, hence 0% probability that 10 of the tosses will fall heads and 10 will fall tails.
Item c:
For the approximation, the mean and the standard deviation are:
[tex]\mu = np = 20(0.5) = 10[/tex]
[tex]\sigma = \sqrt{np(1 - p)} = \sqrt{20(0.5)(0.5)} = \sqrt{5}[/tex]
Using continuity correction, this probability is [tex]P(10 - 0.5 \leq X \leq 10 + 0.5) = P(9.5 \leq X \leq 10.5)[/tex], which is the p-value of Z when X = 10.5 subtracted by the p-value of Z when X = 9.5.
X = 10.5:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{10.5 - 10}{\sqrt{5}}[/tex]
[tex]Z = 0.22[/tex]
[tex]Z = 0.22[/tex] has a p-value of 0.5871.
X = 9.5:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{9.5 - 10}{\sqrt{5}}[/tex]
[tex]Z = -0.22[/tex]
[tex]Z = -0.22[/tex] has a p-value of 0.4129.
0.5871 - 0.4129 = 0.1742.
0.1742 = 17.42% probability that 10 of the tosses will fall heads and 10 will fall tails.
A similar problem is given at https://brainly.com/question/24261244
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