For the binomial distribution with the given values for n and p, state whether o
ID: 3207786 • Letter: F
Question
For the binomial distribution with the given values for n and p, state whether or not it is suitable to use the normal distribution as an approximation. n = 31 and p = .9
Normal approximation is not suitable.
Normal approximation is suitable.
.2946
.3229
.1871
.4936
Normal approximation is not suitable.
Normal approximation is suitable
.2852
.7852
.2148
.7611
0.9318
0.0424
0.1739
0.7248
0.0934
0.0823
0.0869
0.9066
Normal approximation is not suitable.
Normal approximation is suitable.
Use the normal distribution to approximate the desired probability. Find the probability of getting at most 30 fives in 200 tosses of a fair 6-sided die..2946
.3229
.1871
.4936
For the binomial distribution with the given values for n and p, state whether or not it is suitable to use the normal distribution as an approximation. n = 54 and p = .7Normal approximation is not suitable.
Normal approximation is suitable
A library determines that its books are returned on time 90% of the time. If 450 books are checked out, what is the probability that more than 400 will be returned on time?.2852
.7852
.2148
.7611
Assume that women's heights are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. If 90 women are randomly selected, find the probability that they have a mean height between 62.9 inches and 64.0 inches.0.9318
0.0424
0.1739
0.7248
Two percent of CD-ROM drives produced in a certain plant are defective. Estimate the probability that of 10,000 randomly selected drives, at least 219 are defective.0.0934
0.0823
0.0869
0.9066
Explanation / Answer
here n(1-p) =31*0.1=3.1 which is less then 5; hence not suitable
for die : probability of gettting a five =1/6
hence mean =np=200/6 =33.33
and std deviation =(np(1-p))1/2 =5.27
hence probabilty of gettting at most 30 five, out of 200 =P(X<=30)=P(Z<(30.5-33.33)/5.27)=P(Z<-0.53)=0.2852
For the binomial distribution with the given values for n and p, state whether or not it is suitable to use the normal distribution as an approximation. n = 54 and p = .7
normal approximation is suitable as np>5 and n(1-p)>5
3) mean =63.6 and std deviation =2.5
hence std error of mean =2.5/(90)1/2=0.2635
P(62.9<X<64)=P((62.9-63.6)/0.2635<Z<(64-63.6)/0.2635)=P(-2.6563<Z<1.5179)=0.9355-0.0040=0.9315
hence option 0.9318 is right
4) p=0.02
mean =np=1000=0.02*10000=200
std deviation =(np(1-p))1/2 =14
hence P(X>219)=1-P(X<219)=1-P(Z<(218.5-200)/14)=1-P(Z<(1.3214)=1-0.9068=0.0931
option 0.0934 is correct.