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Tutors Answer Your Questions about Describing-distributions-with-numbers (FREE)
Question 1039303: Find out arithmetic mean from the following data by step deviation method:
Marks |Frequency
12.5-17.5 | 4
17.5-22.5 | 20
22.5-27.5 | 17
27.5-32.5 | 15
32.5-37.5 | 2
37.5-42.5 | 5
42.5-47.5 | 5
47.5-52.5 | 2
Click here to see answer by robertb(5830)   |
Question 1040609: Which formula do you use to find the mean (Central Limit Theorem) I am not sure how to calculate this. Can you show the steps to the calculation?
Thanks
It is known that in the absence of treatment, 72% of the patients with a certain illness will improve. The Central Limit Theorem tells us that the percentages of patients in groups of 500 that improve in the absence of treatment are approximately normally distributed. Find the mean of the normal distribution given by the Central Limit Theorem.
A)72%
B)59%
C)34%
D)68%
Click here to see answer by rothauserc(4718)  |
Question 1042139: I need help on parts A and B please.
6. The Acme board of directors has 14 members. In how many ways can one choose…
a) a committee of 3 members (President, VP, Treasurer)? _____________________________
b) a delegation of 3 members where all have equal standing?____________________________
Click here to see answer by solver91311(24713)  |
Question 1042200: Please help me on this i am confused on parts A through F. This is all one problem.
2. Using the following frequency table for a class of Beginning Algebra students’ test scores,
exam score 70 74 78 82 86 90 96 100
frequency 6 6 4 1 6 5 5 2
calculate the following:
a) average _________
b) 25th percentile _________
c) median _________
d) 75th percentile _________
e) 40th percentile _________
f) 80th percentile _________
Click here to see answer by stanbon(75887) |
Question 1069681: the mean score of the students who took a mathematics test was 6. exactly 60% of the student pass the test. the mean score of the student who passed the test was 8. what is the mean score of the student who failed the test?
Click here to see answer by KMST(5328)  |
Question 1074944: Suppose a simple random sample of size n = 36 is taken from a population with mean = 64 and standard deviation = 18.
a) What are the mean and standard deviation of the sampling distribution?
b) What is P(x < 62.6)?
I generally can figure this out if given the formulas but I don't know the formulas in this instance.
Click here to see answer by Boreal(15235)  |
Question 1077764: The income (in Rands) of seven drivers during the week are 1080,2000,1580,1540,2500,1800,1580. Calculate standard deviation of income can you please correct me from the answer I have
1080+2000+1580+1540+2500+1800+1580=12080/7=1725.71
1080-1725.71=(-645.71)^2 =416941.40
2000-1725.71=(274.29)^2 =75235
1580-1725.71=(-145.71)^2 =21231.40
1540-1725.71=(-185.71)^2 =34488.40
2500-1725.71=(774.29)^2=599525
1800-1725.71=(74.29)^2=5519
1580-1725.71=(-145.71)^2=21231.40
S^2=1174171.60/7
= 167738.80
S=√167738.8
= 409.56
Therefore standard deviation is 409.56 please correct me
Click here to see answer by Boreal(15235) |
Question 1077813: Income (in Rands) of drivers during the week are 1080,2000,1540,1580,2500,1800,1580 find Quartile Deviation
Please correct me from my answering
Sol: step 1 rearrangement of numerical order
1080,1540,1580,1580,1800,2000,2500
Let x equal to number of count
x=7
x=7/2=3.5
1580 is the third value in Q1
1540 is the second value in Q1
Q1=(1540+1580)/2= 1560
Q1=1560
2000 is the second value in Q3
2500 is the third value in Q3
Q3=(2000+2500)/2=2250
Q3 = 2250
QD= (Q3-Q1)/2
= (2250-1560)/2
= 345
Therefore quartile deviation is 345 correct me if I am wrong
Click here to see answer by Boreal(15235) |
Question 1089133: The 12 monthly averages of daily high temperatures for San Antonio are given in the data set that follows. The high temperature on April 1 was 45° F. What was the z-score for that day?
48.5, 52.5, 63.2, 68.9, 73.9, 81.4, 85.6, 85.1, 78.8, 70.7, 59.6, 51.0
a. –1.75
b. –1.86
c. 1.75
Click here to see answer by jim_thompson5910(35256) |
Question 1102088: student receives test scores of 52, 83, and 91. The student final exam is 88 and homework score is 76. Each test is worth 20% the final exam is worth 25%, and the homework grade is worth 15%. What is the students mean score in the class?
Mean:
Standard deviation:
Click here to see answer by Theo(13342)  |
Question 1107735: A statistical analysis of 1000 long distance calls made from the headquarter of Microsoft Corporations indicates that length of these calls is normally distributed with mean 240 seconds and standard deviations 50 seconds. What is the probability that a call lasted less than 185 seconds?
Click here to see answer by rothauserc(4718) |
Question 1113411: Suppose x has a distribution with μ = 30 and σ = 25.
(a) If a random sample of size n = 39 is drawn, find μx, σ x and P(30 ≤ x ≤ 32). (Round σx to two decimal places and the probability to four decimal places.)
I do not know how to solve for P(30≤ x ≤ 32), an explanation would be awesome!
Cheers
Click here to see answer by rothauserc(4718) |
Question 1115785: A box has three cards numbered
1
,
2
, and
3
A bag has three balls labeled
A
,
B
, and
C
Juan will randomly pick a card from the box and record the number chosen.
Then he will randomly pick a ball from the bag and record the letter chosen.
Give the sample space describing all possible outcomes.
Then give all of the outcomes for the event that the letter chosen is
C
.
Use the format
1A
to mean that the number chosen is
1
and the letter chosen is
A
.
If there is more than one element in the set, separate them with commas.
Samplespace:
EventthattheletterchosenisC:
Click here to see answer by stanbon(75887) |
Question 1118707: The mean salary of 5 employees is $34600. The median is $35700. The mode is $36900. If the median paid employee gets a $3900 raise, then ...
What is the new mean?
What is the new median?
What is the new mode?
Click here to see answer by Boreal(15235) |
Question 1124718: Let x be a random variable that represents the weights in kilograms (kg) of healthy adult female deer (does) in December in a national park. Then x has a distribution that is approximately normal with mean μ = 63.0 kg and standard deviation σ = 7.9 kg. Suppose a doe that weighs less than 54 kg is considered undernourished.
(a) What is the probability that a single doe captured (weighed and released) at random in December is undernourished? (Round your answer to four decimal places.)
(b) If the park has about 2850 does, what number do you expect to be undernourished in December? (Round your answer to the nearest whole number.)
does
(c) To estimate the health of the December doe population, park rangers use the rule that the average weight of n = 55 does should be more than 60 kg. If the average weight is less than 60 kg, it is thought that the entire population of does might be undernourished. What is the probability that the average weight
x
for a random sample of 55 does is less than 60 kg (assuming a healthy population)? (Round your answer to four decimal places.)
(d) Compute the probability that
x
< 64.3 kg for 55 does (assume a healthy population). (Round your answer to four decimal places.)
Suppose park rangers captured, weighed, and released 55 does in December, and the average weight was
x
= 64.3 kg. Do you think the doe population is undernourished or not? Explain.
Since the sample average is below the mean, it is quite unlikely that the doe population is undernourished.
Since the sample average is above the mean, it is quite unlikely that the doe population is undernourished.
Since the sample average is below the mean, it is quite likely that the doe population is undernourished.
Since the sample average is above the mean, it is quite likely that the doe population is undernourished.
Click here to see answer by ikleyn(52790)  |
Question 1132499: a certain student approach you to ask for help to organize, present and interpret data he gathered. you are to use statistical tools for a specific type of data and make correct and accurate interpretation.
example of data are; production of crops, data of grades, bills of electricity in different years and others.
please compute the data and interpret it using concepts of distributions and z-scores.
Click here to see answer by ikleyn(52790) |
Question 1135757: A farmer has found that the length of string beans he grows follows a normal distribution, having a mean of 12 cm, and a standard deviation of 1.17 cm. If the farmer can only sell beans between 9 and 15 cm long, about what percentage of the crop cannot be sold?
Click here to see answer by Boreal(15235) |
Question 1136578: A psychologist is interested in the conditions that affect the number of dreams per month that people report in which they are alone. In the general population the number of such dreams per month follows a normal curve, with μ = 5 and σ = 4. The psychologist studies 36 individuals and found that their mean number of alone dreams is 8. What is the Z score for this groups mean in the distribution of sample means?
please & thank you
Click here to see answer by Theo(13342) |
Question 1136580: A sample of 36 dieters are in a weight lost program. People who do this program typically loss 40 lbs of weight (with a standard deviation of 12). This sample of dieters lost 45 lbs. Obtain the z score for this sample.
Click here to see answer by Theo(13342) |
Question 1136572: 1) A population has a mean of μ = 50 and a standard deviation of σ= 12.
For samples of size n = 16, what is
the mean (expected value) and
the standard deviation (standard error) for the distribution of sample means?
I found the standard deviation (standard error) = 3
but for the mean would it just be 50?
Thank you in advance.
Click here to see answer by Theo(13342) |
Question 1142277: The mean salary of 5 employees is $40300. The median is $38500. The lowest paid employee's salary is $32000. If the lowest paid employee gets a $3100 raise, then ...
a) What is the new mean?
New Mean = $
b) What is the new median?
New Median = $
Click here to see answer by Boreal(15235) |
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