Question

A supervisor has determined that the average salary of the employees in his department is $40000 with a standard deviation of $15000. A sample of 25 of the employees salaries was selected at random. Assuming the distribution of the salaries is normal, what is the probability that the average for this sample is between $36000 and $42000?

Answer 1

Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

Answer 2

I have no clue how to solve this problem.

Answer 3

\[\bar{X}～N(40000,(\frac{15000^2}{25})\]\[\frac{15000^2}{25}=9\times10^6\]\[P(\bar{X}<42000)=P(Z<\frac{42000-40000}{\sqrt{9\times10^6}})\approx P(Z<0.67)=\Phi(0.67)=0.7486\]Therefore :\[P(\mu<\bar{X}<42000)\approx0.7486-0.5=0.2486\]\[P(36000<\bar{X}<\mu)=P(\bar{X}<44000)-0.5=P(Z<\frac{44000-400000}{\sqrt{9\times10^6}})-0.5\]\[\approx P(Z<1.33)-0.5\approx \Phi(1.33)-0.5=0.9082-0.5=0.4082\]So:\[P(36000<\bar{X}<42000)\approx 0.2486+0.4082=0.6568\]

Answer 4

Thank you so much! :)

Answer 5

God, that took a long time! :P

Answer 6

haha ty i have been staring at that problem for 2 days.......dont go anywhere i got another one for ya lol.
im aaron btw

Answer 7

I hope it's nothing that takes as long as an estimation question, I'm tired now :P

Answer 8

lol i know right

Answer 9

By the way, I'm sorry if a mistake in that :P I'm guessing you haven't seen one if you haven't said.

Answer 10

A tire manufacturer claims that the lives of its tires are normally distributed and will last an average of 60,000 miles with a standard deviation of 3000 miles. 64 randomly selected tires were tested and the average miles where the tire failed were recorded. What is the probability that the mean miles recorded when the tires failed will be more than 59,500 miles?

Answer 11

it was correct lol

Answer 12

How do you know? You have the answers?

Answer 13

Yes it tells me if I have it worked out correctly.

Answer 14

the new one looks just as confusing lol

Answer 15

Arghh, it just deleted all my work but basically, I get 0.9082.

Answer 16

lol i hate when that happens

Answer 17

Do you live in America?

Answer 18

that was correct, lol ok one more problem you have just literally saved me 10 hours im terrible at this..........yes i live in st. louis mo

Answer 19

Ok, I'm assuming your in ～12th grade, yeah? I strongly recommend this book: http://www.amazon.com/Statistics-AQA-SMP-AS-Mathematics/dp/052160527X/ref=sr_1_1?ie=UTF8&s=books&qid=1307222760&sr=8-1

Answer 20

The average life of a particular battery is determined to be 56.2 hours. With a standard deviation of 4.3 hours. Assume battery life is normally distributed. Batteries in the lower 10% do not meet minimum life requirements.

What is the minimum life requirement for a battery?

Answer 21

Damn! I accidentally pressed refresh and it deleted all my work again. 50.69?

Answer 22

yeppers, can you help me on a few more? i would appreciate it

Answer 23

Ok, I could do with practising these anyway :)

Answer 24

Ages of applicants for a particular scholarship are normally distributed with a mean age of 21.3 years and a standard deviation of 1.85 years. If an applicant is randomly selected, find the probability that the applicant is under 18 years old.

Answer 25

still here?

Answer 26

\[X～N(21.3,1.85^2)\implies P(X<18)=1-P(X<24.6)=1-P(Z<\frac{24.6-21.3}{1.85})\]\[\approx1-P(Z<1.78)=1-\Phi(1.78)=1-0.9625=0.0375\]

Answer 27

sweet rock on dont go anywhere lol

Answer 28

so the final answer was .04?

Answer 29

Yeah, but normally, in statistics, we give more significant figures than that.

Answer 30

ok thats cool were supposed to round up to 2 decimals

Answer 31

Oh, that's somewhat ridiculous but whatever.

Answer 32

lol yeah i know

Answer 33

Song lengths scheduled for airplay on a radio station ar normally distributed with a mean length of 3.5 minutes and a standard deviation of .28 minute. What percentage of songs is within 3 to 4 minutes long?

Answer 34

This is making so much more sense I appreciate it.

Answer 35

That makes it worthwhile.

Answer 36

\[X～(3.5,0.28)\implies P(3<X<\mu)=P(\mu<X<4)=P(X<4)-0.5\]
\[=P(Z<\frac{4-3.5}{0.28})-0.5\approx P(Z<1.79)-0.5=\Phi(1.79)-0.4\]\[=0.9633-0.5=0.4633\]\[\implies P(3<X<4)=2\times0.4633=0.9266\]

Answer 37

Can I go to bed now?

Answer 38

one more please lol

Answer 39

Go on then.

Answer 40

3.) The average life of a particular bacteria strain is determined to be 27.3 hours with a standard deviation of 2.6 hours. Assume bacteria life is normally distributted. Bacteria in the lower 10% are determined to be inconsequential.l

What is the minimum life required for studay?

Answer 41

23.97 hrs

Answer 42

i actually have one more if you dont mind staying realy quick

Answer 43

i over looked it

Answer 44

Ok, lets see it.

Answer 45

k just a sec :)

Answer 46

A catering company claims that the total calories of their entrees are normally distributed with an average of 1000 calories and a standard deviation of 300 calories. Fifty randomly selected meals were tested and the average calories were recorded.

What is the probability that the mean calories recorded is greater than 1100 calories?

Answer 47

there you go :)

Answer 48

\[X～N(1000,300^2)\implies \bar{X}～N(1000, \frac{300^2}{50})\Longleftrightarrow \bar{X}～N(1000, 1800)\]\[\implies P(\bar{X}>1100)=1-P(\bar{X}<1100)=1-P(Z<\frac{1100-1000}{\sqrt{1800}})\]\[\approx P(Z<2.36)=\Phi(Z)=0.9909\]

Answer 49

hey are you usually on here? thanks for your help!

Answer 50

Damn forgot to do 1-ans, it should be 1-0.9909=0.0091

Answer 51

ok :)

Answer 52

Yeah, I'm normally on here, feel free to ask me any time :)

Answer 53

ok great thank you! :)

Answer 54

That's fine.

Answer 55

hello are you here today?

Answer 56

hi

Answer 57

Hi.

Answer 58

Hello are you here?

Answer 59

Just post a question and I'll answer it when I'm next online :)

Answer 60

1,) A university surveys 36 randomly selected students and asks the number of dollars spent on textbooks/supplies during the academic year. The results are (in dollars are shown below.


425 78 53 699 663 282
845 373 161 22 853 653
366 806 920 562 828 931
641 134 916 38 379 90
388 894 654 639 916 284
700 684 600 425 717 822

Construct a 90% confidence interval for the mean dollars spent.

Would a given mean value of $600 be likely, given the calculated
confidence level?

Answer 61

aaron.vallero@yahoo.com is my email

Answer 62

Hi are you here?

Answer 63

Yep.

Answer 64

For that confidence interval, can you work out the mean for me? And do you know what the unbiased estimator S is?

Answer 65

i believe the mean is the $600 but thats all i understand with it

Answer 66

Do you know how to calculate the mean?

Answer 67

I thought it was the sum of all the numbers

Answer 68

The mean is the sum of all the numbers divided by the amount of numbers there are. S is what we use to estimate the standard deviation of a population from a sample, it is given by:\[S=\frac{\sum_{i=1}^{n}x _{i}^2-n \bar{x}^2}{n-1}\]Are you ok with this notation?

Answer 69

Yeah thats fine

Answer 70

Ok, so now we've estimated the standard deviation, \(\bar{X}\) follows a t distribution with n-1 (which in this case is 35) degrees of freedom. If we then look up 95% in the percentage points of the t distribution with 35 degrees of freedom, this will give us the t value we need to use, which is 1.69 (use 95% because you want 5% each side). So the interval is:\[\bar{x}\pm 1.69\frac{\S}{\sqrt{n}}\approx540\pm 1.69\frac{294.96}{\sqrt{36}}\approx540\pm83.1\]Using the proper notation:\[(456.9,623.1)\]

Answer 71

$600 is in the interval and therefore, seems possible. Oh, I made one mistake, what I said was \(S\) was actually \(S^2\) so you need to take the square root to get \(S\). Please say if any of this is unclear.

Answer 72

hey are you here?

Answer 73

??