Question

In a plantation of sunflowers, the height of plants tends to follow a normal distribution, with a mean of 2.47 m and a standard deviation of 0.28 m. a) Calculate, to the nearest %, the probability that a randomly-selected sunflower from this plantation measures 2.47 m in height, using a measuring stick that is precise to the nearest cm? b) Calculate, to the nearest %, the proportion of sunflowers from this plantation that are taller than 300 cm in height. c) Calculate, to the nearest %, the probability that a randomly-selected sunflower from this plantation has a height between 200 and 300 cm. d) It has been decided that the shortest 10% of sunflowers in the plantation are to be harvested for floral arrangements. Calculate, to the nearest cm, the maximum height for a sunflower to be harvested for this purpose. e) A certain species of crawling insect is known to favour the roots of the variety of sunflower grown in this plantation. If 10 of these insects arrive in the plantation and each one selects a different sunflower, and it is assumed that they are unable to discern the heights of the plants from their position on the ground, then what is the probability, to the nearest %, that fewer than 3/4 of these insects will choose a sunflower with a height between 200 and 300 cm?

Answer 1

a)

µ =    2.47          
σ =    0.28          
              
P( X ≤    2.47   ) = P( (X-µ)/σ ≤ (2.47-2.47) /0.28)      
=P(Z ≤   0.00   ) =   0.5000 or 50% (answer)

b)

µ =    2.47                  
σ =    0.28     

X=300cm =3 m   
                      
P ( X > 3   ) = P( (X-µ)/σ ≥ (3-2.47) / 0.28)              
= P(Z > 1.89   ) = P( Z <   -1.893   ) =    0.0292 or 2.92% (answer)

c)

µ =    2.47                                  
σ =    0.28                                  
we need to calculate probability for ,                                      
P (   2   < X <   3   )                      
=P( (2-2.47)/0.28 < (X-µ)/σ < (3-2.47)/0.28 )                                      
                                      
P (    -1.679   < Z <    1.893   )                       
= P ( Z <    1.893   ) - P ( Z <   -1.68   ) =    0.9708   -    0.0466   =    0.9242 or 92.42% (answer)

d)

µ=   2.47                  
σ =    0.28                  
P(X≤x) =   0.1000                  
                      
Z value at    0.1   =   -1.2816   (excel formula =NORMSINV(   0.1   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   -1.282   *   0.28   +   2.47  
X   =   2.11 m

  

the maximum height for a sunflower to be harvested for this purpose = 211cm

e)

P(2m

n=10

p=0.9242

P ( X = 0) = C (10,0) * 0.9242^0 * ( 1 - 0.9242)^10=      0.0000
P ( X = 1) = C (10,1) * 0.9242^1 * ( 1 - 0.9242)^9=      0.0000
P ( X = 2) = C (10,2) * 0.9242^2 * ( 1 - 0.9242)^8=      0.0000
P ( X = 3) = C (10,3) * 0.9242^3 * ( 1 - 0.9242)^7=      0.0000
P ( X = 4) = C (10,4) * 0.9242^4 * ( 1 - 0.9242)^6=      0.0000
P ( X = 5) = C (10,5) * 0.9242^5 * ( 1 - 0.9242)^5=      0.0004
P ( X = 6) = C (10,6) * 0.9242^6 * ( 1 - 0.9242)^4=      0.0043
P ( X = 7) = C (10,7) * 0.9242^7 * ( 1 - 0.9242)^3=      0.0301

P(fewer than 3/4) = P(X<3/4*10) = P(X<7.5) = P(X≤7) =ΣC(n,x)*px*(1-p)(n-x) = 0.0349