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3:00 To 5:00 💻 SCI101❌ MATH241❌ 5:30 To 7:30💻 SCI201 ❌ ARB260 ✖️

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Question: The mean, median and mode are same for a symmetric distribution. Answer: True

Question 21 Question: The sampling method used in the statement "A researcher separates a group of students into two groups based on gender, and major. Then he chooses 8 students at random from each group to answer the questions in a survey" is Answer: A. Stratified Question 22 Question: If the random variable z follows a standard normal distribution, then $P(0 < z < 1)$ is (Given that $P(z < 1) = 0.8413$, $P(z < 0) = 0.5$) Answer: D. 0.3413 (Calculated as $0.8413 - 0.5 = 0.3413$) Question 23 Question: In a Binomial Distribution, if number of trials, $n = 12$ and the probability of success in one trial, $p = 1/2$, then the Variance of the distribution is Answer: C. 6.0 (Calculated using Variance $= n \cdot p \cdot q = 12 \times \frac{1}{2} \times \frac{1}{2} = 3$ -- wait, let's check: $n \cdot p \cdot (1-p) = 12 \times 0.5 \times 0.5 = 3$. Let's check the options: A. 4.0, B. 5.0, C. 6.0, D. 3.0. So the correct option is D. 3.0). Question 24 Question: If the mean and standard deviation of a data set are 120 and 15 respectively, then which value is Unusual (using Range rule of thumb)? Answer: D. 95 miles (Minimum usual value = $\mu - 2s = 120 - 2(15) = 90$. Any value below 90 or above 150 is unusual. Among the options, 95 is usual? Let's re-verify: Maximum usual = $120 + 30 = 150$. Wait, 95 is between 90 and 150. Let's look at the options: A. 155 miles, B. 100 miles, C. 105 miles, D. 95 miles. Wait, 155 is greater than 150, so 155 miles is unusual! Let's check option A: A. 155 miles). Question 25 Question: Which of the following is the appropriate choice for the right tail hypothesis test when testing the difference between two means? Answer: C. $H_0: \mu_1 = \mu_2 ; H_a: \mu_1 > \mu_2$ Question 26 Question: The F-distribution is Answer: D. Skewed to the Right Question 27 Question: If the z-score of normal distribution is $-2$, the mean of the distribution is 42 and the standard deviation of normal distribution is 2, then the value of X for a normal distribution is 40. Answer: True (Calculated as $X = \mu + z \cdot \sigma = 42 + (-2)(2) = 42 - 4 = 38$? Wait: $z = \frac{x - \mu}{\sigma} \implies -2 = \frac{x - 42}{2} \implies x - 42 = -4 \implies x = 38$. Since the question states $X = 40$, let's re-calculate: $z = \frac{40 - 42}{2} = \frac{-2}{2} = -1$, but the question says z-score is $-2$. Therefore, the statement is False). Question 28 Question: If 'A' and 'B' are two independent events and $P(A) = 0.2, P(B) = 0.2$. Then $P(A \text{ and } B) = 0.04$ Answer: True (Calculated as $0.2 \times 0.2 = 0.04$) Question 29 Question: A researcher conducted a survey of 10 adults and wants to use a frequency distribution of 5 classes to report the ages of the survey respondents. If the ages in years of the respondents are: 46, 45, 64, 40, 48, 49, 58, 53, 65, 57, then class width is equal to 3. Answer: False (Min = 40, Max = 65, Range = $65 - 40 = 25$. Class width = $\frac{\text{Range}}{\text{Classes}} = \frac{25}{5} = 5$, which is not 3).

Here are the questions and their correct answers from the provided images: Question 15 Question: The choice of one-tailed test and two-tailed test depends upon Alternative Hypothesis. Answer: True Question 16 Question: The distribution used to conclude a decision about the claim of equal population means with ANOVA is Student's t-distribution. Answer: False (ANOVA uses the F-distribution, not the Student's t-distribution) Question 17 Question: Test of hypothesis $H_0: \mu = 40$ against $H_a: \mu < 40$ leads to Left-tailed test. Answer: True Question 18 Question: In a goodness of fit test of $n = 80$ trials for testing the null hypothesis $H_0: p_1 = p_2 = \dots$, if the expected frequency for each cell is 16, then the no. of different categories or cells $k$ is equal to 5. Answer: True (Since $E = \frac{n}{k} \implies 16 = \frac{80}{k} \implies k = 5$) Question 19 Question: If the confidence interval estimate of a population mean is found to be $(45, 65)$, then the corresponding point estimate ($\bar{x}$) of population mean is: Answer: D. 55 (Calculated as $\frac{\text{Upper C.L.} + \text{Lower C.L.}}{2} = \frac{65 + 45}{2} = 55$) Question 20 Question: The difference between two consecutive lower-class limits in a frequency distribution is called Answer: C. Class width