Registered Sec#____ Name_____________Thursday Sec#______ TAMUSTAT#______Homework #61. As we keep tossing a coin (as n increases), which of the following happens?A. The sample proportion, p gets smaller.No, the sample proportion p, will flucuate about the true proportion, , so it could get smaller OR larger (assuming it is a random sample). Although, p = x/n, as n increases, so does x, so p’s size stays relatively the same.B. The sample proportion, p gets closer to , the population proportion.Yes, the larger the sample the closer the estimate, p, is to the true parameter value, .C. The standard deviation of the sample proportion, p, gets smaller.Yes, p = (1)/n, so as n increases, p decreases. D. All of the above will happen. No, obviously.*E. Only two of the above will happen. Yes, B and C.2. Why use (or report) an average of several observations instead of just one?A. You could have made a mistake with one, but it's less likely you'd make the same mistake with several.No. Although this might be a true statement, it is not the reason we use the mean. Notice, you could be taking biased observations, so the mean would also be biased (you’d be making the same mistake).B. The average is less biased than any individual observation.No, both the means, x’s, and the observations, x’s, are unbiased (they’re mean is ).C. An average can't be an outlier. No, a sample mean can be an outlier, but it will still be closer to  than some of the x’s. Notice the outlier in both of the boxplots for the sample means in HW3, #2.*D. Averages are less variable than the individual observations.Yes, (x) = (x)/n, so the sample means will always be les variable than the x’s (unless n = 1, then it would be the same).E. An individual observation can't represent the mean of a whole population.No, since the x’s are unbiased, any of them COULD represent the population (they just wouldn’t be very ‘good’ representatives).3. Assuming the graph above represents the probability distribution for X, what is P (0.5 < X < 1.0)?P(0.5 < X < 1.0) = 0.5*(1.0  0.5)*0.5 = 0.125 (1/2)*base*heightA. 0.5*B. 0.125C. 0.25D. 0.375E. 0.625 4. For the same distribution, what is P (X = 0.5)?X is a CONTINUOUS variable; therefore, the probability thatIt EQUALS anything is ALWAYS 0!*A. 0B. 0.375C. 0.5D. 0.25E. 0.1255. What is the 79th percentile for the standard normal, Z ~N(0, 12)?P(Z < z*) = 0.79, looking up 0.79 in the body of the table, you’ll find z* = 0.81A. 0.79B. 0.7852C. 0.2148*D. 0.81Registered Sec#____ Name_____________Thursday Sec#______ TAMUSTAT#______E. -0.816. Let X ~ N(25, 42). What is P (20 < X < 26)?P(20 < X < 26) = P((2025)/4 < Z < (2625)/4) = P(1.25 < Z < 0.25) = P(Z < 0.25)  P(Z < 1.25) = 0.59870.1056 = 0.4931*A. 0.4931B. 0.7043C. 0.4013D. 0.8413E. 0.6853 7. We've talked alot about `describing a distribution'. What does this mean?A. It means we need to know the 5 Number Summary.No, this gives us an idea of the shape, center and spread, but we need the actual distribution(e.g., normal or uniform),  and .B. It means we need to know that the data is normal. No, this is not enough.C. It means we need to know the shape, maximum and minimum. No, the minimum and maximum don’t give us , plus we also need the mean, *D. It means we need the shape, center and spread.Yes, this gives us the distribution(shape), mean, , (center) and standard deviation, , (spread) so we can find probabilities.E. It means we need to know the mean and the median since we can then tell if the distribution is symmetric or skewed left or right. No, we still need actual distribution and the standard devation.8. Let 9X~ N (5, 22). What is the range of the middle 90% of these9X? What are xa* and xb* such that P(xa* <9X < xb*) = 0.90 (centered at the mean,  = 5)?We must find the z*’s first, and convert to9X. P(z* < Z < z*) = 0.90  P(Z < z*) = 0.95 and P(Z < z*) = 0.05  z* = 1.645xa* =   z* = 5  (1.645)2 = 1.71 and xb* =  + z* = 5 + (1.645)2 = 8.29*A. (1.71, 8.29)B. (-1.645, 1.645)These are the z*’s, but we need to convert to 9XC. (-8.29, 8.29)This is not centered at the mean, 5.D. (-1.28, 1.28)This is not centered at the mean, 5.E. (2.44, 7.56)This is centered at the mean, 5, but it is too wide.9. If I had asked for the middle 95% instead, whichof the following would be true?A. The interval would be wider since the standard deviation would be larger.No, changing the percentage only changes the z*’s, not the standard deviation.B. The interval would be narrower since the standard deviation would be smaller. No, changing the percentage only changes the z*’s, not the standard deviation.*C. The interval would be wider since it covers more of the possible observations. Yes, increasing the percentage increases the z*’s, therefore making the interval wider.D. The interval would be narrower since it'smore accurate.No, the interval would be wider.E. The interval would be the same since the mean, , and standard deviation, , would not change.No, the z*’s change.10. When is a sample size of 30 not enough to say the distribution of approximately normal?A. when the data is categorical and the true proportion of successes, is less than 15%Registered Sec#____ Name_____________Thursday Sec#______ TAMUSTAT#______Yes, the rule for categorical data is n and n(1)  5. If  = 0.15, n = 30*0.15 = 4.5, which is< 5.B. when the data is already normalNo, 30 is actual more than what we need. When the data is normal, the sample mean is also normal, now matter what n is used.C. when the data is highly skewedYes, for highly skewed data (extremely NONnormal), 30 is not enough for the sample means to be consistently normally distributed.D. All of the above are true statements. No*E. Exactly two of the above are true statements. 11. Let p42 ~ N(0.7, 0.0712). What is P (p42 < 0.5)?P(p42 < 0.5) = P(Z < (0.50.7)/0.071) = P(Z < 2.82) = 0.0024A. 0.5B. 0C. 0.9976*D. 0.0024E. 0.212. Why do we call the