Refer to the following information for Questions 6 and 7 Con

Refer to the following information for Questions 6 and 7.

Consider selecting one card at a time from a 52-card deck. (Note: There are 4 aces in a deck of cards)

6. If the card selection is with replacement, what is the probability that the first card is an ace and the second card is also an ace? (Express the answer in simplest fraction form)

7. If the card selection is without replacement, what is the probability that the first card is an ace and the second card is also an ace? (Express the answer in simplest fraction form)

True or False.

1. (a) If the variance of a data set is zero, then all the observations in this data set are zero.

(b) If P(A) = 0.4 , P(B) = 0.5, and A and B are disjoint, then P(A AND B) = 0.9.

(c) Assume X follows a continuous distribution which is symmetric about 0. If , then .

(d) A 95% confidence interval is wider than a 90% confidence interval of the same parameter.

(e) In a right-tailed test, the value of the test statistic is 1.5. If we know the test statistic follows a Student’s t-distribution with P(T < 1.5) = 0.96, then we fail to reject the null hypothesis at 0.05 level of significance .

Please show all work work and give justification for #1. Please no handwritten work.

Solution

6.

There are 4 aces and cards are drawn with replacement. So the probability of drawing an ace remain same in both draws. So the probability that the first card is an ace and the second card is also an ace is (4/52)*(4/52)=1/169.

7.

Now cards are drwan without replacement so the probability that first card is an ace is 4/52. After selecting first card, there are 3 ace remaining in remaining 51 cards. So the probability that secodn card is an ace is 3/51. So the probability that the first card is an ace and the second card is also an ace is (4/52)*(3/51)=1/221