Question

In Exercises 9-15, find the indicated probability using the standard normal distribution. 9. Pz 1.28) 10. P(z>-0.74) 11. P(-2.15-г<1.55) 1 2. P(0.42 3.5) 13. Pz-2.50 or z>2.50) 14. P(Z<O or z>1.68) 15. Yearly amounts of black carbon emissions from cars in India are normally distributed, with a mean of 14.7 gigagrams per year and a standard deviation of 11.5 gigagrams per year. Find the probability that the amount of black carbon emissions from cars in India for a randomly selected year are (a) less than I2.3 gigagrams per year. (b) between 15.4 and 19.6 gigagrams per year. (c) greater than 17.7 gigagrams per year. (Adapted from Atmospheric Chemistry and Physics)

Answer 1

Note:

$X\sim N(\mu, \sigma^2)\Rightarrow Z=\frac{X-\mu}{\sigma} \sim N(0,1)\\\\ \Phi(z)= cdf \: of \: Z= P(Z\leq z)$

We use R Studio or standard normal table to find the cdf, in R studio syntax is pnorm(z)

9)

$P(Z<1.28)= \Phi(1.28)=0.8897274$

10)

$P(Z>-0.74)=1-P(Z<0.74)=1- \Phi(0.74)=1-0.77035 = 0.22965$

11)

$P(-2.15<Z<1.55)= P(Z<1.55)- P(Z<-2.15)= \Phi(1.55)-\Phi(-2.15)=0.9236516$

12)

$P(0.42<Z<3.15)= P(Z<3.15)- P(Z<-0.42)= \Phi(3.15)-\Phi(0.42)=0.3364264$

13)

Using inclusion exclusion principle

P(Z<-2.5 or Z>2.5) = P(Z<-2.5) + P(Z>2.5) -P(Z<-2.5 and Z>2.5)

=P(Z<-2.5) + P(Z>2.5) - 0 ( there is no common point between Z<-2.5 and z>2.5)

  $= P(Z<-2.5)&plus;1- P(Z<2.5)\\\\=\Phi(-2.5)&plus;1-\Phi(2.5)\\\\ =0.006209665 &plus;1-0.9937903 \\\\\ =0.01241933$

14)

Using inclusion exclusion principle

P(Z<0or Z>1.68) = P(Z<0) + P(Z>1.68) -P(Z<0 and Z>1.68)

=P(Z<0) + P(Z>1.68) - 0 ( there is no common point between Z<0 and z>1.68)

$= P(Z<0)&plus;1- P(Z<1.68)\\\\=\Phi(0)&plus;1-\Phi(1.68)\\\\ =0.5 &plus;1-0.9535213 \\\\\ =0.5464787$

15) X= amount of black carbon emissions from cars in india.

X~N( 14.7, 11.5² )

$Z=\frac{X-14.7}{11.5} \sim N(0,1)$

a) P(X<12.3)

$P(X<12.3)= P(\frac{X-14.7}{11.5}<\frac{12.3-14.7}{11.5})\\\\ =P(Z<-0.2087)\\\\ =\Phi(-0.2087)\\\\ =0.5826588$

b) P(15.4<X<19.6)

$=P(X<19.6)- P(X<15.4)\\\\ = P(\frac{X-14.7}{11.5}<\frac{19.6-14.7}{11.5})-P(\frac{X-14.7}{11.5}<\frac{15.4-14.7}{11.5})\\\\ =P(Z<0.4261)-P(Z<0.0609)\\\\ =\Phi(0.4261)-\Phi(0.0609)\\\\ =0.1407019$

c)P(X>17.7)

= 1-P(X<17,7)

$= 1-P(\frac{X-14.7}{11.5}<\frac{17.7-14.7}{11.5})\\\\ =1-P(Z<0.2609)\\\\ =1-\Phi(0.2609)\\\\ =1-0.6029152\\\\ = 0.3970848$

R COde:

pnorm(1.28)
pnorm(0.74)
pnorm(1.55)-pnorm(-2.15)
pnorm(3.15)-pnorm(0.42)
pnorm(-2.5)
pnorm(2.5)
pnorm(0)
pnorm(1.68)
pnorm(0.2087)
pnorm(0.4261)-pnorm(0.0609)
pnorm(0.2609)