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8.1.3 A random sample has been taken from a normal distribution and the following confidence intervals constructed using the sam
1 answer:
Answer:
a) The value of the sample mean is 50
b) (38.02, 61.98) is the 95% as it covers more values than the other.
Step-by-step explanation:
a) Upper and lower values for the confidence intervals are at the same distance from the mean, as they represent an upper and lower limit for the value, and the most probable value is the one in the middle, which, in a normal distribution is the mean.
As they derive from the same data, both pairs of values should give us the same mean:
b) The percentage of the confidence interval tells us which is the area below the bell curve (of the normal distribution) between the limits given.
One could give a value with a 100% CI, but limits should be minus and plus infinity, and as the interval gets smaller, less values are covered (less area), and less certain we are about our value being inside the interval.
Then, we can say that the widest interval must be the 95% CI, i.e. (38.02, 61.98).
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Answers:
(a) Possible values of X are: 0, 1, 2, 3, 4, 5
(b) See attached image below; specifically figure 2
(c) No, it is <u>not</u> symmetric
(d) As a fraction, the answer is 1/6. In decimal form, the answer is approximately 0.1667 (converts to roughly 16.67%)
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Explanations:
(a) The largest possible roll we can get on one die is 6, the smallest is 1, so the largest possible difference is 6-1 = 5. The smallest difference is 0 as we could get doubles (eg: both dice show a "3" at the same time). We can get any difference in between 0 and 5. So X can take on whole numbers from 0 to 5, ie, X can be any number from the set {0, 1, 2, 3, 4, 5}
(b) Consult the attached image. Figure 1 shows a 6 by 6 grid of values that represent all the possible ways to roll two dice. Each value in the interior of the grid is the result of subtracting the outer header values. Figure 2 is where the answer is shown for part b. The idea is that you list out the X values along the top of the table and the probability P(X) values along the bottom row. I show the work on how to compute each probability in the attachment. The probability is equal to the number of occurences over the number total. Example: X = 3 shows up 6 times out of 36, so P(X) is 6/36 = 1/6 when X = 3.
(c) The distribution is not symmetric as figure 3 shows a lop-sided diagram. The mode should be at the center, but this figure shows the mode off to the left side.
(d) Count how many copies of '4' or '5' show up in the interior of the grid shown in figure 1. There are four copies of '4' and two copies of '5'. Overall there are 4+2 = 6 instances of either '4' or '5'. This is out of 36 total, so 6/36 = 1/6 is the fractional form of the answer which approximates to 0.1667 and converts to 16.67% approximately.
