b. The most appropriate data set of the above descriptive statics include low integer value and following the same, small range is covered for the respective student’s payment.
c. It is justified and agreeable term to follow central tendency based on the provided data set.
e. Yes, there are unusual data values
f. Due to the low range and form of SD, it is possible to create a normal population using the above data set.
In to compare students performance: Q3 and Q4 have better performance than Q1 and Q2 and the results for Q3 are nearest of the mean than the results of Q4 (CV of Q3< CV of Q4)
d. The data set is skewed
e. In case of performance, it can be said that all of the students have shown equal performance according to the SD value and mean mode.
Question 4 (Refer to excel)
A = event that alt. 1 fails.
B = event that alt. 2 fails.
a) p(A and B) = p(A)p(B|A), but they are independent, so p(B|A)=p(B)
b) See (a), but notice that p (event that alt. 1 won't fail) = 1-p(A)
c) p(A or B) = p(A)+p(B)-p(A and B)
Let X be the number of alternators that fail. X has the binomial distribution with n = 2 trials and success probability p = 0.02
In general, if X has the binomial distribution with n trials and a success probability of p then
P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)
for values of x = 0, 1, 2, ..., n
P[X = x] = 0 for any other value of x
The likelihood mass capacity is determined by taking a gander at the quantity of blend of x objects that is browsed n questions and after that a sum of x. It is achievement and n - x disappointments or as it were, the binomial is the entirety of n free and indistinguishably appropriated Bernoulli trials
X ~ Binomial( n = 2 , p = 0.02 )
The Probability Mass Function, PMF,
f(X) = P(X = x) is:
P( X = 0 ) = 0.9604 <<< Neither fail
P( X = 1 ) = 0.0392 <<< one fails
P( X = 2 ) = 4e-04 <<< Both fail
Question 5 (Refer to Excel)
The 95% confidence interval for the true mean number of steps = [x-bar - t * (s/sqrt n), x-bar + t * (s/sqrt n)] = [3278.72 - 2.1098 * (97.05/sqrt 18), 3278.72 + 2.1098 * (97.05/sqrt 18)] = [3230.46, 3326.98], say [3230 steps, 3327 steps]
Sample size = (t * s/Error)^2 = (2.1098 * 97.05/20)^2 = 104.81, say 105
(c ) The line plot of the data is shown below. It appears that Dave is taking lesser and lesser steps to finish the 2.2-mile course.
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