Mathematics, 23.07.2020 19:01, sindy35111

7.22. (a) A fair coin is tossed 100 times. Estimate the probability that the number of heads is between 40 and 60. Estimate the probability that the number is between 50 and 55.

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answered: amiramoonbeam

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 = $\sqrt{n \times p \times (1-p)}$

Standard deviation σ = $\sqrt{100 \times 0.5 \times (1-0.5)}$

Standard deviation σ = $\sqrt{50 \times (0.5)}$

Standard deviation σ = $\sqrt{25}$

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)

$P(40 < X < 60) = P( \dfrac{X - \mu}{\sigma}< \dfrac{59.5 - 50 }{5}) - P( \dfrac{X - \mu}{\sigma}< \dfrac{39.5 - 50 }{5})$

$P(40 < X < 60) = P( Z < \dfrac{9.5 }{5}) - P( Z< \dfrac{-10.5 }{5})$

P(40 < X < 60) = P( Z

P(40 < X < 60) =0.9713 -0.0178

P(40 < X < 60) =0.9535

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)

$P(50 < X < 55) = P( \dfrac{X - \mu}{\sigma}< \dfrac{54.5 - 50 }{5}) - P( \dfrac{X - \mu}{\sigma}< \dfrac{49.5 - 50 }{5})$

$P(50 < X < 55) = P( Z < \dfrac{4.5 }{5}) - P( Z< \dfrac{-0.5 }{5})$

P(50 < X < 55) = P( Z

P(50 < X < 55) =0.8159 -0.4602

P(50 < X < 55) =0.3557

Therefore; the probability that the number of heads is between 50 and 55 is 0.3557

answered: Guest

15%

step-by-step explanation:

use the table and count the number of guesses that has two correct answers.

the following are two correct answers

10010, 01100, 11000

a total of 3 numbers having 2 correct guesses out of 20 questions.

probability of getting 2 correct guesses = the number of questions getting 2 correct answers / total number of questions.

= 3/20

to convert it into percent multiply by 100, we get

= 3/20 * 100

= 15%

the answer is 15%

you.

Afive question multiple-choice quiz has five choices for each answer. use the random number table pr

answered: Guest

boxplot e is the answer

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7.22. (a) A fair coin is tossed 100 times. Estimate the probability that the number of heads is betw...