Answer with explanation:
Given : The heights of a certain population of corn plants follow a normal distribution with mean [tex]\mu=145\ cm[/tex] and standard deviation [tex]\sigma=22\ cm[/tex]
a) Using formula [tex]z=\dfrac{x-\mu}{\sigma}[/tex], the z-value corresponds to x= 135 will be
[tex]z=\dfrac{135-145}{22}\approx-0.45[/tex]
At x= 155, [tex]z=\dfrac{155-145}{22}\approx0.45[/tex]
The probability that plants are between 135 and 155 cm tall :-
[tex]P(-0.45<z<0.45)=P(z<0.45)-P(z<-0.45)\\\\=0.6736447- 0.3263552\\\\=0.3472895\approx0.3473=34.73\%[/tex]
Hence, 34.73% of the plants are between 135 and 155 cm tall.
b) Sample size : n= 16
Using formula [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex], the z-value corresponds to x= 135 will be
[tex]z=\dfrac{135-145}{22}{\sqrt{16}}\approx-1.82[/tex]
At x= 155, [tex]z=\dfrac{155-145}{22}{\sqrt{16}}\approx1.82[/tex]
The probability that plants are between 135 and 155 cm tall :-
[tex]P(-1.82<z<1.82)=P(z<1.82)-P(z<-1.82)\\\\= 0.9656205- 0.0343795\\\\=0.931241\approx0.9312=93.12\%[/tex]
Hence,The percentage of the samples would the sample mean height be between 135 and 155 cm.= 93.12%
