Suppose A is a positive real number and mA is the average value of (sin(Ax))3 on the
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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
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
.
- Fair coin, hence
.
The probability is <u>P(X = 10)</u>, thus:
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 and standard deviation
, the z-score of a measure X is given by:
- 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 <u>x successes on n trials, with p probability</u> of a success on each trial. It can be approximated to the normal distribution with
.
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:
Using continuity correction, this probability is , which is the <u>p-value of Z when X = 10.5 subtracted by the p-value of Z when X = 9.5.</u>
X = 10.5:
has a p-value of 0.5871.
X = 9.5:
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 brainly.com/question/24261244