1. True or False. Justify for full credit. (15 pts)
(a) If the variance of a data set is zero, then all the observations in this data set are zero.
Variance is zero all data are same or zero; hence, given statement is false.
(b) If P(A) = 0.4 , P(B) = 0.5, and A and B are disjoint, then P(A AND B) = 0.9.
If two events are disjoint, then the probability of them both occurring at the same time is 0.
Disjoint: P(A and B) = 0.
Hence, given statement is false.
(c) Assume X follows a continuous distribution which is symmetric about 0. If P(X>2)= 0.3 then
P(X<-3)≤ 0.3
False as for symmetric distribution, P(X<-3)<0.3.
(d) A 95% confidence interval is wider than a 90% confidence interval of the same parameter.
True. We can have it as given below. For 95% confidence interval, z =±1.96; hence, width = 196*2 = 3.92. For 90% confidence interval, z = ±1.645; hence, width = 1.645*2 = 3.29; hence, given statement is true.
(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 .
Here p = 1-0.96 = 0.04 that is less than 0.05. Hence, statement is true.
2.
Check out time / Frequency / Relative frequency1-1.9 / 3 / 3/25=0.12
2-2.9 / 12 / 12/25=0.48
3-3.9 / 5 / 5/25=0.20
4-4.9 / 3 / 3/25=0.12
5-5.9 / 2 / 2/25=0.08
25 / 1
3.(5+3+2)/25 = 0.4 or 40%
4. N=25; hence, median will occur at (25+1)/2=13th position and it is in 2-2.9 class interval.
5.left skewed
6. Probability that first card is ace = 4/52
Probability that 2nd card will be ace = 3/51 as card is not replaced.
Hence, probability that both cards will be ace = (4/52)*(3/51) = 1/221
7. Probability that first card is ace = 4/52
Probability that 2nd card will be ace = 4/52 as card is replaced.
Hence, probability that both cards will be ace = (4/52)*(4/52) = 1/169
8. For quiz 1, interquartile range = 85-45 = 40
For quiz 2, interquartile range = 90-35 = 55
Hence, quiz 1 has less interquartile range.
9. It is impossible to tell using only the given information.
10. It is impossible to tell using only the given information.
11. We can draw the following figure to reply these questions.
Number of students having laptop = 650+150 = 800
Number of students having tablet = 150+150 = 300
Number of students having laptop as well as tablet = 150
Number of having nothing = 1000-(650+150+150) = 50
Hence, probability that randomly selected student will have nothing = 50/1000 = 0.05
12. Probability that a randomly selected student has a laptop, given that he/she has a tablet
= 150/300
= 0.5
13. These are not independent events as some students have both laptops as well as tablets.
14. There are 50 numbers from 0 to 49.
Number of ways selecting 3 numbers without repeating = 50P3= 50*49*48 = 117600
15. (a)We can have 1, 2 or 3 heads. The following table gives probability distribution.
The possibilities are as follows.
HHH 3 heads
HHT 2 heads
HTH 2 heads
HTT 1 head
THH 2 heads
THT 1 head
TTH 1head
TTT no head
Let X is the number of heads in three tosses of a coin.
X / PROBABILITY0 / 1/8
1 / 3/8
2 / 3/8
3 / 1/8
(b) Mean of x = 0*(1/8)+1*(3/8)+2*(3/8)+3*(1/8) = 1.5
Standard deviation =
=
=1.1456