Question 1: Suppose the true probability mass function for the number of days un

Question

Question 1: Suppose the true probability mass function for the number of days until discharge from the hospital for premature babies is known to be characterized as follows: + Probability fY) 0.14J 0.26 0.35^ 0.19H 0.06 Days until Discharge (Y)* a) What are the true mean, median, mode, and variance for this discrete distribution? b) Are the properties of a probability mass f unction satisfied? Explain c) Determine the following probabilities +-' Prob(Y2) +' a. c. Prob(Y>4) *' d) Suppose two premature babies are born on a particular day, one at 5:00 a.m. and the other at 7:30 a.m. Assuming that their responses are independent, - a. what is the probability that the 5:00 a.m. baby will be discharged in 2 days AND the 7:30 am baby will be discharged in 1 day b. what is the probability that one of the two babies will be discharged in 2 days AND the other baby will be discharged in 1 day

Explanation / Answer

from above given data:

a) true mean =2.77

median =value below and above which 50% values lies =3

mode =3 (as maximum probability occurs at y=3)

variance =1.1971

b) Yes ; as each individual probability lies from 0 to 1 and their sum =1.

c) a)P(Y<2) =P(Y=1) =0.14

b)P(3<Y<=5) =P(Y=4)+P(Y=5) =0.19+0.06=0.25

c)P(Y>4) =P(Y=5) =0.06

d)

a) probability =P(Y=2)*P(Y=1)=0.26*0.14 =0.0364

b)probability =P(first baby in 2 day and second in 1 day+first in 1day and second in 2 days)

=0.26*0.14+0.14*0.26=0.0728

y p(y) yP(y) y2P(y) (y-)2 (y-)2P(y) 1 0.1400 0.140 0.140 3.133 0.439 2 0.2600 0.520 1.040 0.593 0.154 3 0.3500 1.050 3.150 0.053 0.019 4 0.1900 0.760 3.040 1.513 0.287 5 0.0600 0.300 1.500 4.973 0.298 total 1 = 2.77 8.870 10.265 2= 1.1971 std deviation=     =    2 = 1.0941

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