Respuesta :
Answer:
the probability that the number of heads is between 40 and 60 is 0.9535
the probability that the number of heads is between 50 and 55 is 0.3557
Step-by-step explanation:
From the given information:
A fair coin is tossed 100 times.
Let consider n to be the number of time the coin is tossed, So n = 100 times
In a fair toss of a coin; the probability of getting a head P(Head) = 1/2 = 0.5
If we assume X to be the random variable which follows a binomial distribution of n and p; therefore , the mean and the standard deviation can be calculated as follows:
Mean μ = n × p
Mean μ = 100 × 1/2
Mean μ = 100 × 0.5
Mean μ = 50
Standard deviation σ = [tex]\sqrt{n \times p \times (1-p)}[/tex]
Standard deviation σ = [tex]\sqrt{100 \times 0.5 \times (1-0.5)}[/tex]
Standard deviation σ = [tex]\sqrt{50 \times (0.5)}[/tex]
Standard deviation σ = [tex]\sqrt{25}[/tex]
Standard deviation σ = 5
Now, we've made it easier now to estimate the probability that the number of heads is between 40 and 60 and the probability that the number is between 50 and 55.
To start with the probability that the number of heads is between 40 and 60 ; we have:
P(40 < X < 60) = P(X < 60)- P(X < 40)
Applying the central limit theorem , for X is 40 which lies around 39.5 and 40.5 and X is 60 which is around 59.5 and 60.5 but the inequality signifies less than sign ;
Then
P(40 < X < 60) = P(X < 59.5) - P(X < 39.5)
[tex]P(40 < X < 60) = P( \dfrac{X - \mu}{\sigma}< \dfrac{59.5 - 50 }{5}) - P( \dfrac{X - \mu}{\sigma}< \dfrac{39.5 - 50 }{5})[/tex]
[tex]P(40 < X < 60) = P( Z < \dfrac{9.5 }{5}) - P( Z< \dfrac{-10.5 }{5})[/tex]
[tex]P(40 < X < 60) = P( Z <1.9}) - P( Z< -2.1)[/tex]
[tex]P(40 < X < 60) =0.9713 -0.0178[/tex]
[tex]P(40 < X < 60) =0.9535[/tex]
Therefore; the probability that the number of heads is between 40 and 60 is 0.9535
To estimate the probability that the number is between 50 and 55.
P(50 < X < 55) = P(X < 55)- P(X < 50)
Applying the central limit theorem , for X is 50 which lies around 49.5 and 50.5 and X is 55 which is around 54.5 and 55.5 but the inequality signifies less than sign ;
Then
P(50 < X < 55) = P(X < 54.5) - P(X < 49.5)
[tex]P(50 < X < 55) = P( \dfrac{X - \mu}{\sigma}< \dfrac{54.5 - 50 }{5}) - P( \dfrac{X - \mu}{\sigma}< \dfrac{49.5 - 50 }{5})[/tex]
[tex]P(50 < X < 55) = P( Z < \dfrac{4.5 }{5}) - P( Z< \dfrac{-0.5 }{5})[/tex]
[tex]P(50 < X < 55) = P( Z <0.9}) - P( Z< -0.1)[/tex]
[tex]P(50 < X < 55) =0.8159 -0.4602[/tex]
[tex]P(50 < X < 55) =0.3557[/tex]
Therefore; the probability that the number of heads is between 50 and 55 is 0.3557
