Quiz 6.1 – 6.2 tomorrow Test 6.1 – 6.2 on Monday May use both sides of one note card Test 6.1 – 6.2 on Monday

Sections 6.1 and 6.2 Quiz tomorrow

Page 382, E21(a)

Regular die – Tetrahedral die 1 2 3 4 5 6 1 2 3 4

Page 382, E21(a) 1 2 3 4 5 6 1 0 1 2 3 4 5 2 -1 0 1 2 3 4 3 -2 -1 0 1 2 3 4 -3 -2 -1 0 1 2

Page 382, E21(a)

Shape of distribution?

Shape of distribution? Symmetric

Verify this is a probability distribution.

Page 382, E22(a)

Page 382, E22(a) 1 2 3 4 5 6 - 1 2 3 4 5 6 0 0 0 0 0 0

Page 382, E22(a)

An oil exploration firm is to drill ten wells, each in a different location. Each well has a probability of 0.1 of producing oil. It will cost the firm $60,000 to drill each well. A successful well will bring in oil worth $1 million. Taking into account the cost of drilling, what is the firm’s expected gain?

An oil exploration firm is to drill ten wells, each in a different location. Each well has a probability of 0.1 of producing oil. It will cost the firm $60,000 to drill each well. A successful well will bring in oil worth $1 million. a) E(gain) = 1,000,000(10)(0.1) – 60,000(10) = $400,000

Page 379, E1

Page 379, E1 Enter data for “Score” in list 1 and “Proportion of Students” in list 2. Then, use 1-Var Stats L1, L2

Page 379, E1(a) = 2.16 2.2214

Page 379, E1(b)

Page 379, E1(b)

Sample of size 4

Page 379, E1(b) Find the mean score for the sample of size 4.

Page 379, E1(b) Find the mean score for the sample of size 4. 3

Page 392, E30(a)

Page 392, E30(a) P(at least one works correctly) = 1 – P(none work correctly) = 1 – binompdf(2, .92, 0) = 0.9936

Page 392, E30(a) P(at least one works correctly) = 1 – P(none work correctly) = 1 – binompdf(2, .92, 0) = 0.9936 or Note: P(failure) is 0.08, so P(at least one works correctly) = 1 – (0.08)2 = 0.9936

Page 392, E30(b)

Page 392, E30(b) P(at least one works correctly) = 1 – P(none work correctly) = 1 – binompdf(3, .92, 0) = 0.9995 or Note: P(failure) is 0.08, so P(at least one works correctly) = 1 – (0.08)3 = 0.9995

What can cause two events to be independent?

What can cause two events to be independent? (1) sampling with replacement (2)

What can cause two events to be independent? (1) sampling with replacement (2) small sample from a large population

What criteria must be met in order to have a probability distribution?

What criteria must be met in order to have a probability distribution? (1) must include all possible outcomes (2) probabilities sum up to 1 (3) no negative probabilities

When is it appropriate to use the normal approximation to estimate the probability for a binomial situation?

When is it appropriate to use the normal approximation to estimate the probability for a binomial situation? Only if both of these are true: np ≥ 10 n(1 – p) ≥ 10

If it is appropriate to use the normal approximation to estimate the probability for a binomial situation how do you determine the expected value and the standard deviation?

If it is appropriate to use the normal approximation to estimate the probability for a binomial situation how do you determine the expected value and the standard deviation? E(X) = μx = np

If it is appropriate to use the normal approximation to estimate the probability for a binomial situation how do you determine the expected value and the standard deviation? E(X) = μx = np σx =

normalcdf(left bound, right bound, mean, standard deviation) If it is appropriate to use the normal approximation to estimate the probability for a binomial situation, then to find the probability we can use: normalcdf(left bound, right bound, mean, standard deviation)

normalcdf(left bound, right bound, mean, standard deviation) If it is appropriate to use the normal approximation to estimate the probability for a binomial situation, then to find the probability we can use: normalcdf(left bound, right bound, mean, standard deviation) normalcdf(left bound, right bound, np, )

Suppose you flip a coin 13 times. (1) what is the probability that you’ll get exactly 7 heads? (2) what is the probability that you’ll get at least 7 heads? (3) what is the probability that you’ll get at most 7 heads?

Suppose you flip a coin 13 times. (1) what is the probability that you’ll get exactly 7 heads? binompdf(13, 0.5, 7) = 0.2095 (2) what is the probability that you’ll get at least 7 heads? (3) what is the probability that you’ll get at most 7 heads?

Suppose you flip a coin 13 times. (2) what is the probability that you’ll get at least 7 heads? 1 – binomcdf(13,0.5, 6) = 0.5 (3) what is the probability that you’ll get at most 7 heads?

Suppose you flip a coin 13 times. (2) what is the probability that you’ll get at least 7 heads? 1 – binomcdf(13,0.5, 6) = 0.5 (3) what is the probability that you’ll get at most 7 heads? binomcdf(13, 0.5, 7) = 0.7095

When you spin a coin, the probability of getting heads is 0. 4 When you spin a coin, the probability of getting heads is 0.4. Suppose you take a random sample of 5 spins. Construct the probability distribution of the random variable, X, defined as the number of heads in your sample.

Number of heads in sample P(x)

Number of heads in sample P(x) 1 2 3 4 5

binompdf(5, 0.4) Number of heads in sample P(x) 0 0.0778 1 0.2592 0 0.0778 1 0.2592 2 0.3456 3 0.2304 4 0.0768 5 0.0102

Questions? Quiz 6.1 – 6.2 tomorrow Test 6.1 – 6.2 on Monday May use both sides of one note card Test 6.1 – 6.2 on Monday

Suppose you select five numbers at random from 10 through 99, with repeats allowed. Find the probability exactly three of the numbers are even.

Suppose you select five numbers at random from 10 through 99, with repeats allowed. Find the probability exactly three of the numbers are even. P(even) = 0.5

Suppose you select five numbers at random from 10 through 99, with repeats allowed. Find the probability exactly three of the numbers are even. P(even) = 0.5 P(X = 3) = binompdf(5, 0.5, 3) = 0.3125

Suppose you select five numbers at random from 10 through 99, with repeats allowed. Find the probability exactly one of the numbers has digits that sum to a number greater than or equal to 9.

Suppose you select five numbers at random from 10 through 99, with repeats allowed. Find the probability exactly one of the numbers has digits that sum to a number greater than or equal to 9. P(sum greater than or equal to 9) = = 0.6

Suppose you select five numbers at random from 10 through 99, with repeats allowed. Find the probability exactly one of the numbers has digits that sum to a number greater than or equal to 9. P(X = 1) = binompdf(5, .6, 1) 0.0768

Suppose you select five numbers at random from 10 through 99, with repeats allowed. Find the probability at least one of the numbers has digits that sum to a number greater than or equal to 9.

Suppose you select five numbers at random from 10 through 99, with repeats allowed. Find the probability at least one of the numbers has digits that sum to a number greater than or equal to 9. P(X = at least 1) = 1 – P(X = 0) = 1 – binompdf(5, .6, 0) 0.9898