1. The Length of a certain species of fish has a normal distribution with mean 50 mm and standard deviation 4.5 mm. A random sample of four fish is drawn and their length X1, X2, X3, and X4, are observed. Calculate the probability that.

 (3 pts) All four fish have lengths between 51 and 60 mm
 (3 pts) The mean length of the four fish is between 51 and 60mm
 (3 pts) Explain why the answer to part A would always be bigger than answer to part b.
Answer:
a. P (51 < X < 60) = ( 5150/4.5) < X < ( 60 â€“ 50/4.5)
= 0.22 < X < 2.22
= 0.986791 – 0.587064 = 0.399727
b. Mean of Sampling Distribution = 50 mm
Standard Deviation of the Sample = 4.5/sqrt(4)
= 4.5/2 = 2.25
t (51) = 51 â€“ 50/ (2.25) = 1 / 2.25 = 0.44
t (60) = 60 â€“ 50/ (2.25) = 4.44
Df = 4 1 =3
P (51 < t < 60) = 0.9894 â€“ 0.6551
=0.3343
c. The answer to part A will always be bigger because it is taken from a larger population, whereas the part b is taken from sample, and due to less number of observations, the probability will be less.
2. (3pts) Using computer simulation, several independent random samples each of size n were drawn from a population that has a normal distribution with mean 50 and a standard deviation 9. This means of the samples were then calculated and it was found that 68%of the sample means lies between 48.5 and 51.5. Find the sample size n. Assume that n is large enough for the central limit theorem to hold.
Answer: 48.5 â€“ 50 /( 9/sqrtN) < X < (51.5 50)/ 9 sqrtN) = 0.68
= 1.5/(9/sqrtN)< X< 1.5/(SqrtN) = 0.68
= 1.5/(9/sqrtN) = 0.408
= (1.5* SqrtN) = .408
= SqrtN = 0.408/1.5
Sqrt N = 0.272
N = .073984
3. A random variable x has a binomial distribution with n â€“ 100 and p=.45. Using the Normal approximately to the binomial with continuity correction, calculate the following probabilities
a. P(X=50)
b. P(x>60)
c. P(40<X<50)
a. 0.048151971
b. 0.000938232
c. 0.682367
4. Recall the population of Ellipticus (copy attached). Using stat crunch, drawindependent random samples of 5 nonmutants and 5 mutants from this population. Then measure the lengths (to the nearest millimeter) of Y1, Y2, Y3, Y4, and Y5 be the lengths of the mutants. Using the data
3 points each
A. Obtain a 95 confidence interval for the population mean length Î¼Nof non mutants
B. Obtain a 95 confidence interval for the population means length Î¼M of mutants
C. Obtain a 95 confidence interval for the difference in population mean length Î¼NÎ¼Mbetween nonmutants and mutants
5. The manufacturer of a new prostate cancer drug claims that it has a least 65% cure rate. You are the statistician assigned to validate the claim with data. So you administered the drug on a random sample of 30 patients and proceed with the following test:
Let p be the unknown cure rate and let the random variable x be the number of cures among the 30 patients.note that x has a binomial distribution with n30 and event probablity p.
Set up the null hypotheses ho: p0.65
Decision rule: reject ho if the observed number of cures x>25
3 point each
A. Using statcrunch calculate y(p) for p= .1,.2,.3,.4,.5,.6,.7,.8,.9
B. Graph y(p) versus p using excel (line graph)
C. Dicuss briefly how good you this test is based on they(p) above.
Answer:
The the line is straight sloped line, it shows the there is a constant increase in success with increase in the p or sample
D. Construct a test x> a for the null hypothesis ho: p=0.65 that will allow you to 10% chance of making a type 1 error (i.e. rejecting a true h0) with maximum probability of rejecting ho y (0.8) when in fact the true state of nature is ha: p=0.8
Test will be based on the changes, with the rejection region of 10% and same the chance of making type I error. The test will consist the sample, with the probability of success change to 0.85