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We strive to offer the best statistics, accounting, finance, economics and chemistry homework help online.
We strive to offer the best statistics, accounting, finance, economics and chemistry homework help online.

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Question: Suppose a normal distribution has a mean of 20 and a standard deviation of 4.

What is the z-score of a value that is 0.52 standard deviations less than the mean?

Answer: x = 20 - 0.52*4 =17.92

z = (17.92 - 20)/4 = -0.52 (Ans)
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Question: A laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 193 milligrams with s = 18.7 milligrams. Construct a 95 percent confidence interval for the true mean cholesterol content of all such eggs.
a. 181.2 < < 204.8
b. 181.0 < < 205.0
c. 181.1 < < 204.9
d. 183.3 < < 202.7

x-bar = 193
sd = 18.7
t_c = critical at 11 df and alpha=0.05 is = 2.201

margin of error, E = (18.7*2.201)/sqrt(12) = 11.8814

95% CI = (x-bar - E, x-bar + E )
= (193 - 11.8814, 193 + 11.8814)
= (181.1186, 204.8814)

c. 181.1 < x-bar < 204.9 (Ans.)
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Question: A bag contains n cards, each having one of the consecutive numbers written on it, with each number being used once. The probability of drawing a card with a number less than or equal to 10 is 4/10.How many cards are in the bag?

Answer: P(x < = 10) = 4/10

Or, P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) + P(x = 5) + P(x = 6) + P(x = 7) + P(x = 8) + P(x = 9) + P(x =10) = 4/10

or, 1/n + 1/n + 1/n + 1/n + 1/n + 1/n + 1/n + 1/n + 1/n + 1/n = 4/10
or, 10/n = 4/10
or, n = 10*10/4 = 25 (Ans.)
Question: If the probability of frozen pipes is 0.63, find the probability of 8 or more frozen pipes out of the eleven.?

p = 0.63
P(x >= 8) = P(x = 8) + P(x = 9) + P(x = 10) + P(x=11)
= 11c8*(0.63)^8*(1-0.63)^3 + 11c9*(0.63)^9*(1-0.63)^2 + 11c10*(0.63)^10*(1-0.63)^1
+ 11c11*(0.63)^11*(1-0.63)^0
= 0.3714 (Ans.)
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Question: If five standard coinsare tossed find the probability of all 5 coins landing on tails?

Answer: sample space = 2^5 = 32

The required probability = 1/32 (Ans.)
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Question: We want to estimate the difference between the mean starting salaries for recent graduates with mechanical engineering and electrical engineering degrees from the University of Michigan (UM). The following information is available: A random sample of 59 starting salaries for UM mechanical engineering graduates produced a sample of \$48,736 and a standard deviation of \$4,430. A random sample of 30 starting salaries for UM electrical engineering graduates produced a sample mean of \$53,208 and a standard deviation of \$4,286. Please derive a 95 %confidence interval for the difference between the true mean salaries.

(Hint: We do not know the population standard deviations but the sample sizes are large enough (≥ 30). Hence, you can use sample standard deviations as approxi- mations to population standard deviations)

n1 = 59
x-bar1 = 48,736
s1 = 4,430
----------------------
n2 = 30
x-bar2 = 53,208
s2 = 4,286

z_c = critical z = 1.96 at 95% confidence.
margin of error, E = sqrt(s1^2/n1 + s2^2/n2)*1.96
= sqrt(4430^2/59 + 4286^2/30)*1.96
= 1905.29

95% CI = (x-bar1 - x-bar2 - E , x-bar1 - x-bar2 + E)
= (48736 - 53208 - 1905.29 , 48736 - 53208 + 1905.29)
= (-6377.29 , -2566.71) (Ans.)
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Question: Maria and Zoe are taking Biology 101 but are in different classes. Maria’s class has an average of 78% with a standard deviation of 5% on the midterm while Zoe’s class has an average of 83% with a standard deviation of 12%. Assume that scores in both classes follow a normal distribution. Now suppose that Maria scored an 84 on the midterm exam and Zoe scored an 89.

Convert Maria’s midterm score of 84 to a standard z-score, rounded to the nearest tenth, and place it in the blank. z-score =

Answer: z = (84 - 78)/5 = 1.2
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Question: A can contains 500 grams of cat food. The chart below(in the link) shows that the weight is normally distributed around the average of 490 grams and standard deviation is 5 grams.
Now a department store buys 3000 cans. How many of these cans can be expected to contain at least the 500g cat food as stated on the lable?

Answer: z = (500 - 490)/5 = 2

P(x > = 500) = P(z > = 2) = 0.0228

No. of cans expected to contain at least the 500g cat food as stated on the lable
= 3000*0.0228 = 68.4 = 69 (Ans.)
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Question: I am calculating a 95% confidence interval for the mean number of advertisements in a magazine. There is a total of 112 pages and 54 advertisements. The mean and proportion I have calculated is 0.48 (I believe they should be the same, correct me if I'm wrong). Is there a way to do this using the calculator?

p = 54/112 = 0.4821
z_c = 1.96 at 95% confidence.

margin of error, E = sqrt(pq/n)*z_c
= sqrt(0.4821*(1- 0.4821)/112)*1.96
= 0.0925

95% CI = (p - E , p + E) = (0.4821 - 0.0925 , 0.4821 + 0.0925)
= (0.3896, 0.5746) (Ans.)
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Question: There is a new drug that is used to treat leukemia. The following data represents the remission time in weeks for a random sample of 21 patients using the drug.

10 7 32 23 22 6 16
11 20 19 6 17 35 6
10 34 32 25 13 9 6

Let X be a random variable representing the remission time in weeks for all patients using the new drug. Assume that the distribution of x is normal. A previously used drug treatment has a mean remission time of 12.5 weeks. Does the data indicate that the mean remission time using the new drug is different from 12.5 week at a level of significance of 0.01?

State the Null Hypothesis
x ≥ 12.5 ?

State the Alternatice Hypothesis
x ≤ 12.5 ?

State the Level of significance

State the Test Statistic

Perform Calculations

Statistical Conclusion

Experimental Conclusion

Ha : mu not = 12.5

level of significance = 0.01

Descriptive Statistics
Column 1

Sample Size, n: 21
Mean: 17.09524
Median: 16
Variance, s^2: 99.99048
St Dev, s: 9.999524

test statistics, z = (17.1 - 12.5)/(10/sqrt(21)) = 2.1079

critical z = +/- 2.58 at 0.01 level.

Hence, computed z (2.1079) is within (-2.58, 2.58).

Fail to reject Ho.

conclusion : the mean remission time using the new drug is not different from 12.5 week