This question has three parts. (a) Consider a large population with defect ratio

Question

This question has three parts. (a) Consider a large population with defect ratio of 1%. We draw a large number of samples, each of size 25 and count the number of defective items in each sample. Let denote the number of defective items found in sample i, i = 1, 2, N, where N > 1. Estimate the median, first quartile and third quartile of the sequence D,, D (b) Let X ~ N(-2, 2-0.04). Consider a large number of observations of X, that give the values x1,..,"N, where N » 1 . Estimate the 12th percentile of the observations. (c) Consider the Fibonacci sequence given by the recurrence relation F, Fn-1 + Fn-2, F(0) = F(1) = 1. Write the Fibonacci sequence up to n you wrote is the number 21? 15. What percentile of the set of numbers

Explanation / Answer

a)

Expected number of defectives in each sample=mean=0.01*25=0.25

standard dev=sqrt(0.01*25*(1-0.01))=sqrt(0.25*0.99)=0.5

median:

P(Z<=z)=0.5

z=0

z=(x-mean)/std dev

0=(x-0.25)/0.5

x=0.25

median=0.25

First quartile:

P(Z<=z)=0.25

z=-0.6745

-0.6745=(x-0.25)/0.5

x=0.25-0.5*0.6745=0.25-0.34=-0.09

First quratile can not be negative,so it will be 0

Third Quartile:

P(Z<=z)=0.75

z=0.6745

x=0.25+0.5*0.6745=0.25+0.34=0.59

third quartile is 0.59

b)

To compute the 12th percentile, we use the formula X= + Z, and we will use the standard normal distribution table.

hence to see in the standard normal distribution table to find the area under the curve closest to 0.12, and from this we can determine the corresponding Z score. Once we have this we can use the equation X= + Z, because we already know that the mean and standard deviation are 2 and 0.2, resp.

When we go to the table, we find that the value 0.12 is not there exactly, however, the values 0.1210 and 0.1190 are there and correspond to Z values of -1.17 and -1.18, resp

Using Z=-1.17 the 12th percentile distribution for mean is: X = 2 - 1.17(0.2) = 1.766

and this is the 12 th percentile.

c)

Fn = Fn1 + Fn2

F(0)=1 , F(1)=1,

F(2)=F(1)+F(0)=1+1=2 , F(3)=F(2)+F(1)=2+1=3 ,F(4)=F(3)+F(2)=5 ,F(5)=F(4)+F(3)=8 and similarly

F(6)=8+5=13, F(7)=13+8=21, F(8)=34,F(9)=55,F(10)=89,F(11)=144,F(12)=233,F(13)=377,F(14)=610,F(15)=987

there are total 16 observation (0 to 16) and 21 is 8th observation so its percentile=100*(7+0.5*1)/16=46.875

so 47 th percentile

Pecentile=100*(CF+0.5F)/n, CF=cumulative frequency, F=frequency, n=sample size

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.