Answer: The correct answer is -2,5/2 which is the last answer.
Step-by-step explanation:
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Answer:
The distribution is
b) skewed.
The sum of the probabilities is:
1
Step-by-step explanation:
In a binomial distribution, p represents the probability of success. Success in the sense that the event of interest happens. In the model presented, the probability of success p is 0.4 since we are informed that 40% of adults watch a particular television show.
The next quantity of significance in a binomial model is the number of independent trials, n. In our case there are 6 independent trials since we are told that 6 adults were selected at random. If we let the random variable K denote the number of adults out of the 6 who watch the television show, then K is a binomial random variable with parameters;
n = 6 and p = 0.4
A binomial distribution is only symmetric when either p is 0.5 or n is large. In the presented scenario none of this conditions is met since p is 0.4 while n is just 6 which is relatively small. Thus we conclude that the distribution is not symmetric but rather skewed.
The sum of the probabilities is any discrete probability distribution such as the bernoulli, binomial, negative binomial, poisson, or the geometric distribution is always equal to 1. That's a rule of thumb.
Answer:
a) 29.23% probability that a randomly selected home run was hit to right field
b) 29.23% probability that a randomly selected home run was hit to right field, which is not lower than 5% nor it is higher than 95%. So it was not unusual for this player to hit a home run to right field.
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes. It is said to be unusual if it is lower than 5% or higher than 95%.
(a) What is the probability that a randomly selected home run was hit to right field?
Desired outcomes:
19 home runs hit to right field
Total outcomes:
65 home runs
19/65 = 0.2923
29.23% probability that a randomly selected home run was hit to right field
(b) Was it unusual for this player to hit a home run to right field?
29.23% probability that a randomly selected home run was hit to right field, which is not lower than 5% nor it is higher than 95%. So it was not unusual for this player to hit a home run to right field.
