Question

Question 8 З рі (a) Population :e-44"/x!, x > 0 (b)u =? (c)o? =? (d)n = 9 (e)ā value = a = 4.5 (f) sem = (g)z = (h)Compute P(3< X<a) Step 1: What is your sem? (Rounded to four decimal places entered as ***** Step 2: What is your z score for a value only, rounded to two decimal places? (enter it as ***) Step 3: What is your answer to part (h), required probability, rounded to four decimal places, entered as * ***

Answer 1

In this case we have the probability mass function of the poisson distribution. Therefore by making use of the poisson distribution the mean and Variance can be obtained and hence the required probability that the sample mean of 9 observations lies between 3 and 4.5 can easily be obtained.

- Q8 (a) Population: ēl you inzo. Let X be a random variable which follows poisson distribution, then its probability mass function is given by, PCX = ") — ē y qué t me ta! 2= 0,1,2,-..cs Then, — у (6) Mean I ll = 2=4 (c) Variance = 02 = 2=4 In case of Poisson distribution Mean and variance both are equal and is equal to 1. d) n=9 e u value= a= 4.5. f) sem = o g) - ñ-ll sem h) Compute P( 32 Xsa).

Step 1 - What is your sem? We know that sem = o. Since o= V Variance = J4 = d. As Variance = y, also n=9. Thus, sem r - O-C6cco Therefore, sem for n=9 is 0.6666. Step 2-8 Z-score for a value only as Z = X-el sem n=2= We have 4.5 and l=4 . Trias, Z = ă-l sem 4.5-4 0.5 -= 0.75000 0.6666 - 0.6666 = Thes, ascore is 0.75oon 0.75 rep 3 - In this case, we have to obtain P(32 X £ a) sie, P(32X = 4.5).

We have that - P( 3 2 À 54.5) = ? het Xq, X2, - - .gXg be random sample of size 9. Then Trix E (5) - FC EXO) = 1 (xi) Var (8) = Varl 2x) = varlxi] [ koulun dependent]. = box SX 4 = of = Juance) = 1 / 2 = 2 = 0.666 In order to obtain the Probability of P (32 Xe 4.5), we make use of normal distribution under asymptotes. Therefore, P(35x24.5) = 1 *) X-E(X), YE E(X)

Therefore, the probability is 0.7066 that the X values lies between 3 and 4.5. cs
Sem value is 0.6666 and the value of z score is 0.75.