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Answer:

Approximately, arithmetic mean equals to #32.4#.

Explanation:

First, the theory behind the method.

If a random variable #xi# takes values #x_1#, #x_2#, ...#x_n# with corresponding probabilities #p_1#, #p_2#, ...#p_n#, its Mathematical Expectation or mean, by definition, equals to
#E(xi) = x_1*p_1+x_2*p_2+...+x_n*p_n = Sigma_(i in [1;n])(x_i*p_i)#

Assume, numbers #x_i# are large enough to make this calculation inconvenient. We can always transform this formula using two constants #a# and #h# freely chosen to our liking into (assuming summation #Sigma# is performed for all #i# from #1# to #n#):
#E(xi) = Sigma (x_i*p_i) = h*Sigma ((x_i-a)/h*p_i) + Sigma (a*p_i) = #
#= h*Sigma ((x_i-a)/h*p_i) + a*Sigma (p_i) =#
#= h*Sigma ((x_i-a)/h*p_i) + a#
(since #Sigma (p_i) = 1#)

Now it's up to us to choose constant #a# and #h# in such a way that simplifies the calculations as much as possible.

If values #x_i# that our random variable #xi# takes are distributed with equal intervals (steps), the best results are achieved if #a# is chosen approximately in the middle of these numbers and #h# is the step.

For example, if values #x_i# are #10000, 20000, 30000, 40000, 50000#, choosing #a=30000# and #h=10000# results in #(x_i-a)/h# to be #-2,-1,0,1,2#, which is a much easier to deal with than the original very large numbers.

Addressing our problem, we will have values #x_i# chosen as midpoint of each interval:
interval #[12.5-17.5]# has midpoint at #15#
interval #[17.5-22.5]# has midpoint at #20#
interval #[22.5-27.5]# has midpoint at #25#
interval #[27.5-32.5]# has midpoint at #30#
interval #[32.5-37.5]# has midpoint at #35#
interval #[37.5-42.5]# has midpoint at #40#
interval #[42.5-47.5]# has midpoint at #45#
interval #[47.5-52.5]# has midpoint at #50#

For number #a# we can choose #30# since it's somewhere in the middle of a group of interval midpoints, for number #h# we can choose #5# since it is, obviously, an increment from one value to another.

Probabilities our random variable takes the above values are approximated by real frequencies of taking these values. Each such frequency is a ratio of the number of times this value occurred (#4# for the first interval, #20# - for the second etc.) divided by the total number of experiments #N=4+20+17+15+2+5+5+2=70#.

Now the mean value of our random variable is evaluated as
#E(xi) = 5*((15-30)/5*4/70+(20-30)/5*20/70+#
#+(25-30)/5*17/70+(30-30)/5*15/70+#
#+(35-30)/5*2/70+(40-30)/5*5/70+#
#+(45-30)/5*5/70+(50-30)/5*2/70) + 30 =#
#= 5/70*((-3)*4+(-2)*20+(-1)*17+0*15+#
#+1*2+2*5+3*5+4*2) + 30 = #
#5/70*(-12-40-17+0+2+10+15+8)+30=#
#=1/14*(-34)+30=32.428571...#

In this case it seems sufficient to approximate the mean as #32.4#.

Answer:

One way to calculate the confidence interval of a sample mean is to use the student's t-distribution to find that with #98%# confidence that the mean lies in the range

#1.55, 3.53#

Explanation:

One way to calculate the confidence interval of a sample mean is to use the student's t-distribution where the limits of the interval are given by:

#m+- t_(1-alpha//2,N-1)s/sqrt(N)#

where

#m# is the sample mean
#alpha# is the residual probability;in our case #alpha = 1-0.98 = 0.02#
#s# is the sample standard deviation
#N# is the number of samples (#N-1# is the number of degrees of freedom)

First we need the sample mean and sample standard deviation:

#m=1/N sum_(i=1)^N x_i=2.54#

and the sample standard deviation:

#s = sqrt(1/(N-1) sum_(i=1)^N(x_i-m)^2) =1.11#

Plugging these values into our first equation we can calculate the limits of the confidence interval for our mean:

#2.54+- t_(0.99,9)1.11/sqrt(10)#

Where we can look up the value of #t_(0.99,9)# in a table or use a function in a software package, like the T.INV function in Microsoft Excel to find the value #2.82#

#2.54+-2.82*1.11/sqrt(10)#

therefore we can say with #98%# confidence that the mean lies in the range

#1.55, 3.53#

Answer:

Answers bellow

Explanation:

a) the population proportion cannot be determined as that represents everyone. The sample, however, allows us to determine a point estimate, #hat p#, which we can assume is close the actual.

The Point estimate is,
#hat p= ("Everyone with a particular characteristic")/("Everyone in the sample")#

or in this case

#hat p= 706/1009#

b) to determine a confidence interval we use the equation,

#hat p-z*sigma < p < hat p+z*sigma#

where z is the z-score for your confidence level and #sigma# is your standard deviation.

For a 95% confidence z= 1.959963985 #~~#1.96

#sigma = sqrt((hat p (1-hatp))/n)#
#sigma = sqrt((706/1009 (1-706/1009))/1009)#

#sigma= 0.014431#

so subbing into,

#hat p-z*sigma < p < hat p+z*sigma#

#706/1009-1.96*0.014431 < p < 706/1009+1.96*0.014431#

#0.6714 < p < 0.728#

c) not sure exactly what you mean.
was think of using this equation but I'm not sure what you want to find out.

#n=(z^2p(1-p))/e^2# with e=0.03

d) a smaller confidence interval 95% #=># 90% contains less values that the actual population could represent (remember that #hat p# is just an estimate of this value)

A 95% confidence interval means that 95% of sample means taken will lie within this range. A 90% confidence interval means that 90% will lie within its range, which is smaller. For example,

95% #=>##0.6714 < p < 0.728#

90% #=>##0.676 < p < 0.7234#

The 90% confidence interval is smaller and thus we are less confident that the actual population mean lies within its range.

Answer:

#"The Reqd. Prob. "= 80/99~~0.8081.#

Explanation:

Let #T# be the event that you have Toothache , and, #C# be the

event that you have Cavity . Then, #C'# denotes the event that

you have No Cavity.

In the Usual Notation of Conditional Probability , we have,

#P(T/C)=0.8=4/5, P(C)=0.05=1/20, and, P(T/(C'))=0.01=1/100,# and, we want, #P(C/T).#

Recall that, #P(A/B)=(P(AnnB))/(P(B))#

#:. P(T/C)=4/5 rArr (P(TnnC))/(P(C))=4/5#

#rArr P(TnnC)=(1/20)(4/5) rArr P(TnnC)=1/25..........(1)#

#"Similarly, "P(T/(C'))={P(TnnC')}/{P(C')}={P(TnnC')}/{1-P(C)}#

#rArr 1/100={P(TnnC')}/(1-1/20) rArr P(TnnC')=(1/100)(19/20), i.e.,#

#P(TnnC')=19/2000..........................(2)#

Now, #(TnnC)nn(TnnC')=Tnn(CnnC')Tnnphi=phi...(ast),# &,

#(TnnC)uu(TnnC')=Tnn(CuuC')=TnnU=T.........(star)#

Thus, from #(ast) & (star)#, we see that, #(TnnC) & (TnnC')#

are mutually exclusive events and #T# is their Union Event.

Therefore, by #(1) & (2),# we get,

#P(T)=P(TnnC)+P(TnnC')=1/25+19/2000=99/2000.#

Finally, the Reqd. Prob. #=P(C/T)={P(CnnT)}/{P(T)}#

#=(1/25)/(99/2000)=80/99~~0.8081 (4dp). #

Enjoy Maths. , and, spread the Joy!

Answer:

#3243/10829#

Explanation:

First let us look at the probability that the first card dealt is an ace and the other four non-aces...

There are #4# aces in #52# cards, so the probability of the first card dealt being an ace is:

#4/52 = 1/13#

The remaining pack consists of #51# cards of which #3# are aces, and #48# not. So the probability of the second card being a non-ace is:

#48/51 = 16/17#

The probability of the third card being a non-ace is:

#47/50#

The probability of the fourth card being a non-ace is:

#46/49#

The probability of the fifth card being a non-ace is:

#45/48 = 15/16#

So the probability of an ace followed by #4# non-aces is:

#1/13*color(red)(cancel(color(black)(16)))/17*47/50*46/49*15/color(red)(cancel(color(black)(16))) = 32430/541450 = 3243/54145#

The other four possible sequences of ace vs non-ace which result in exactly one ace being dealt are mutually exclusive, and will have the same probability as this first case.

So the probability of exactly one ace being dealt is:

#5 * 3243/54145 = 3243/10829#

Answer:

From software: #z=1.554774.#
From table lookup: #z ~~ 1.56.#

Explanation:

If we seek an 88% confidence interval, that means we only want a 12% chance that our interval does not contain the true value. Assuming a two-sided test, that means we want a 6% chance attributed to each tail of the #Z#-distribution. Thus, we seek the #z_(alpha//2)# value of #z_0.06#.

This #z# value at #alpha//2 = 0.06# is the coordinate of the #Z#-curve that has 6% of the distribution's area to its right, and thus 94% of the area to its left. We find this #z#-value by reverse-lookup in a #z#-table.

sixsigmastudyguide.com

Find the closest value in the table to 0.9400 as you can, then see what its row and column is. From observation, we see that 0.9394 and 0.9406 are in the table with #z#-values of 1.55 and 1.56 respectively, and so to err on the side of caution, we'll choose the value that gives us a wider interval, #z=1.56.#

Note: We could also get an answer from software like R, by typing the command #"qt(0.94, Inf)"#, which would give us a more precise value of 1.554774.

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