Question

Consider a binomial distribution with n = 10 trials and the probability of success on a single trial p = 0.85.

(a) Is the distribution skewed left, skewed right, or symmetric?

(b) Compute the expected number of successes in 10 trials.

(c) Given the high probability of success p on a single trial, would you expect P(r ≤ 3) to be very high or very low? Explain.
1) Very low. The expected number of successes in 10 trials is more than 3, and p is so high that it would be common to have so few successes in 10 trials.
2) Very high. The expected number of successes in 10 trials is more than 3, and p is so high that it would be unusual to have so few successes in 10 trials.
3) Very low. The expected number of successes in 10 trials is more than 3, and p is so high that it would be unusual to have so few successes in 10 trials.
4) Very high. The expected number of successes in 10 trials is more than 3, and p is so high that it would be common to have so few successes in 10 trials.

(d) Given the high probability of success p on a single trial, would you expect
P(r ≥ 8) to be very high or very low? Explain.

1) Very low. The expected number of successes in 10 trials is more than 8, and p is so high that it would be unusual to have 8 or more successes in 10 trials.
2) Very low. The expected number of successes in 10 trials is more than 8, and p is so high that it would be common to have 8 or more successes in 10 trials.
3) Very high. The expected number of successes in 10 trials is more than 8, and p is so high that it would be common to have 8 or more successes in 10 trials.
4) Very high. The expected number of successes in 10 trials is more than 8, and p is so high that it would be unusual to have 8 or more successes in 10 trials.

Answer 1

Using the binomial distribution, it is found that:

a) Skewed left.

b) The expected number of successes is of 8.5.

c) 3) Very low. The expected number of successes in 10 trials is more than 3, and p is so high that it would be unusual to have so few successes in 10 trials.

d) 3) Very high. The expected number of successes in 10 trials is more than 8, and p is so high that it would be common to have 8 or more successes in 10 trials.

Binomial probability distribution

Probability of exactly x successes on n repeated trials, with p probability, and has expected value given by:

$E(X) = np$

In this problem, the parameters are: $n = 10, p = 0.85$.

Item a:

• The skewness depends on the value of p.

In this problem, since p  = 0.85 > 0.5, the distribution is skewed left.

Item b:

$E(X) = np = 10(0.85) = 8.5$

The expected number of successes is of 8.5.

Item c:

3 is significantly less than the expected value, hence 3 or fewer of successes would be unexpected, hence option 3 is correct.

Item d:

8 is less than the expected value, hence 8 or more successes are very likely, and option 3 is correct.

To learn more about the binomial distribution, you can take a look at https://brainly.com/question/24863377