Question

Many typical household cleaners contain a toxic chemical. Assume the percent of this toxic chemical in one glass cleaner is approximately normal. The mean percent is 3, the standard deviation of the percent is 1.

1.(4 points) Randomly select one bottle of the glass cleaner, find the probability that the percentage of the toxic chemical is more than 3.255.

2.(4 points) Randomly select 10 bottles of this glass cleaner, find the probability exactly 3 bottles which contain the toxic chemical more than 3.255.

3.(4 points) Find the 80th percentile of the distribution of toxic chemical percent (of the glass cleaner), ?80p80.

4.(4 points) Assume random samples with sample size of 16, find the probability that the sample mean percentage (of the toxic chemical) is less than 2.5.

Answer 1

1)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	3
std deviation   =σ=	1.000

probability that the percentage of the toxic chemical is more than 3.255:

probability =P(X>3.255)=P(Z>(3.255-3)/1)=P(Z>0.26)=1-P(Z<0.26)=1-0.6006=0.3994

(

Note:

if using ti-84 use command :normalcdf(3.255,1000000,3,1)

if using excel use command :1-norm.dist(3.255,3,1,true)

2)

sample size       =n=	10
std error=σx̅=σ/√n=	0.3162

probability =P(X>3.255)=P(Z>(3.255-3)/0.316)=P(Z>0.81)=1-P(Z<0.81)=1-0.79=0.2100

(note:

if using ti-84 use command :normalcdf(3.255,1000000,3,0.3162)

if using excel use command :1-norm.dist(3.255,3,0.3162,true)

3)

for 80th percentile critical value of z=		0.8416
therefore corresponding value=mean+z*std deviation=			3.8416

4)

sample size       =n=	16
std error=σx̅=σ/√n=	0.25000

probability =P(X<2.5)=(Z<(2.5-3)/0.25)=P(Z<-2)=0.0228

if using ti-84 use command :normalcdf(-1000000,2.5,3,0.25)

if using excel use command :norm.dist(2.5,3,0.25,true)