28. Pollen count distribution for Los Angeles in September is not normally distributed, with μ = 8.0, σ = 1.0, n = 64. Find P(x-bar > 9.0).
Is n ≥ 30 or the population is normally distributed? Population is not normally distributed, but n > 30, it's 64 so we can proceed with the problem and use the CLT.
The Central Limit Theorem says:
1) Shape: Approximately normal
2) Center: μ(subscripted x-bar) = μ, meaning the center for both the population and the sample is the same.
Thus μ = 8.0.
3) Spread: σ(subscripted x-bar) = σ/√ n (standard deviation divided by the square root of the sample size)
σ/√ n = 1.0 / √64 = 1/8 = 0.125
Now we draw a picture of a normal distributed centered at 8.0
Then add on what we're interested in: P(x-bar > 9.0).
So we move to the right and mark 9 then shade the area above 9.
Then use the z score formula to find the z score so we can find the probability.
9-8/0.125 = 8.0
A z-score of 8 is so large it's not listed on our table, meaning it's a very unusual observation. So we report the most extreme value we have, 3.49 = .9998
But since we wanted the area to the right, it's 1-0.9998 = .0002
0.02% probability of observing a pollen count greater than 9.0.
31. Boned trout prices are normally distributed, with μ = 3.10, σ = 0.30, n = 16. Find the sample mean price that is smaller than 90% of all sample means.
Is n ≥ 30 or the population is normally distributed? n < 30, but the population is normally distributed, so we can proceed with the problem and use the CLT.
The Central Limit Theorem says:
1) Shape: Normal
2) Center: μ(subscripted x-bar) = μ, meaning the center for both the population and the sample is the same.
Thus μ = 3.10.
3) Spread: σ(subscripted x-bar) = σ/√ n (standard deviation divided by the square root of the sample size)
σ/√ n = .30 / √16 = 0.30/4 = 0.075
Now we draw a picture of a normal distributed centered at 3.10
Then add on what we're interested in: "price that is smaller than 90% of all sample means." We are interested in the price where 90% of all observations are greater than it. Another way of saying this: we're interested in the 10th percentile.
P(x-bar < .1000).
So we move to the left of center and approximate the 10% mark and shade below it.
Now we're ready to use the z-score formula, unfortunately we're missing an observation but we have the percentage we're interested in! So we go to the Z-table and look for the Z value that corresponds to 0.1000 probability. We're looking at a Z-score of -1.28
So plugging into the z score formula:
x-3.1/0.075 = -1.28
x-3.10 = -0.096
x = 3.004
In conclusion, the 10th percentile for trout prices is $3.004 / lb.
51. College Board reports that the mean increase in SAT math scores of students who attend review courses is 18 points. Assume that the standard deviation is 12 points and that the change in score are not normally distributed. We are interested in the probability that the sample mean score increase is negative, indicating a loss of points after coaching. Suppose we take a sample of 40, and we are interested in P(x-bar<0), indicating a loss of points after coaching.
μ = 18
σ = 12
The Central Limit Theorem says:
1) Shape: Approximately normal (40>30)
2) Center: μ(subscripted x-bar) = μ, meaning the center for both the population and the sample is the same.
Thus μ = 18
3) Spread: σ(subscripted x-bar) = σ/√ n (standard deviation divided by the square root of the sample size)
σ/√ n = 12 / √40 = 12/2√10 = 1.897 = 1.90
Now we're ready to use the z-score formula, so plugging into the z score formula:
(0-18) / 1.897 = -9.49
However, the most extreme value on the z-table is -3.49. Finding this probability on the z-table: 0.0002
In conclusion, the probability that a person who received coaching would earn a 0 on the SAT is 0.02%
7.3 #44)
Shaq lead the league with a 58.4% goal percentage.
a) Find the minimum sample size that produces a sampling distribution of p-hat that is approximately normal
When dealing with proportions we have to make sure np ≥ 5 AND n(1-p) ≥ 5. This question is asking us to solve for n algebraically.
a1) n(.584)>5 (dividing both sides by .584)
n> 5/.584 (simplifying)
n> 8.56 (rounding up because you cannot have 8.56 goals)
n>9
a2) n(1-.584)>5
n(.416)>5 (dividing both sides by .416)
n > 5 / .416 (simplifying)
n > 12.02 (rounding up because you cannot having 12.02 goals)
n > 13
In conclusion, the minimum sample size required is 13 because that makes both equations true.
13(.584) > 5 and 13(.416)>5
b) Find μ(subscripted p-hat) and σ(subscripted p-hat) when n = 50.
The Central Limit Theorem says:
1) Shape: Approximately normal [np>5 and n(1-p)>5]
2) Center: μ(subscripted p-hat) = μ, meaning the center for both the population and the sample is the same. Thus μ = 58.4
3) Spread: σ(subscripted p-hat) = √ [p(1-p)/n] (square root of the proportion multiplied by 1 minus the proportion divided by sample size)
√ [.584(1-0.584)/50] = √ [.584(1-0.584)/50] = √ 0.2429/50 = √ .0049 = 0.0697
c)Find the probability that in a sample of 200 shots, Shaq would score more than 120 baskets.
120/200 = .6
Given this proportion we can use the z-score formula
(.6-.584) / .0697 Stop! You can't use that standard deviation it only works for a sample size of 50, we need to recalculate it!
√ [.584(1-0.584)/200] = .0349
Returning to our z-score formula...
(.6-.584) / .0349 = .4591 = .46
Taking .46 to the Z table we find a probability of .6772
However we are interested in the probability he would score MORE than 120 so we want the area to the right, you know what that means! 1-.6772 = 0.3228
Note: this answer varies from the book's, this is because they forgot to recalculate the standard deviation to account of the increased sample size.
8.3 #27)
Find the margin of error E for a 95% confidence interval
a) 5 successes in 10 trials.
5/10 = .5
so p = .5
The margin of error formula is: E = Z α/2 (√ [p-hat(1-(p-hat))/n])
Plugging in what we have
E = Z α/2 (√ [0.5(1-0.5)/10]
E = Z α/2 (0.1581)
For this Z α/2 value we need to look at our confidence interval table provided in table 8.1 on pg. 391, similar to table 8.7 on page 422. (might be a good idea to put those on your note card, just saying...)
So Z α/2 for 95% is 1.96
E = 1.96 (.1581)
E = .3099
8.2 Confidence Intervals for Means
Introducing the T-Distribution: Similar to the Z-Distribution, but has more variability (meaning the data is more spread out)
 |
| Notice: Due to the extra variability there is more info at the extremes. |
Remember: Degrees of Freedom = n-1
Confidence Interval for means = x-bar ± T α/2 (s / √n)
Notice that the standard deviation is similar to the one used for CLT means, but here we have s rather than σ. What does this mean? s is the standard deviation of our sample.
8.2 #6)
We are taking a random sample from a normal population with σ unknown. Find T α/2
a) Confidence level 95%, sample size 10
Degrees of freedom = n-1
Plugging in: 10 -1 = 9
Now we go to the t table and look at the 95% confidence level for a df of 9: 2.262
b) Confidence level 95%, sample size 15
Degrees of freedom = n-1
Plugging in: 15 -1 = 14
Now we go to the t table and look at the 95% confidence level for a df of 14: 2.145
c) Confidence level 95%, sample size 20
Degrees of freedom = n-1
Plugging in: 20 -1 = 19
Now we go to the t table and look at the 95% confidence level for a df of 19: 2.093
13. Confidence level 95%, sample size 25, sample mean 10, sample standard deviation 5.
CI: 95%, n = 25, μ = 10, s = 5.
a) T α/2
df = n-1, 25-1 = 24. Go to the T table find CI for 95% and df = 24, 2.064
b)Margin of Error (E) = T α/2(s/√n)
Plugging in what we know: 2.064 (5/√25) = 2.064 (1) = 2.064
c) Confidence interval for μ given the indicated confidence interval
Lower bound = μ - T α/2(s/√n)
10 - 2.064 = 7.936
Upper bound = μ + T α/2(s/√n)
10 + 2.936 = 12.064
No comments:
Post a Comment