The height of a certain population of corn plants follow a normal distribution w

Question

The height of a certain population of corn plants follow a normal distribution with mean 145 cm and standard deviation 23 cm.

(a) what percentage of the plants are between 135 and 155 cm tall?

(b) suppose we were to choose at a random from the population a large number of samples of 16 plants each. In what percentage of the samples would the sample mean height be between 135 and 155 cm.

(c) if y(bar) represents the mean height of a random sample of 16 plants from the population, what is Pr[135<_Y(bar)<_ 155]

(d) If Y(bar) represents the mean height of a random sample of 36 plants from the population, what is Pr[135<_Y(bar)<_ 155]

Explanation / Answer

Normal Distribution
Mean ( u ) =145
Standard Deviation ( sd )=23
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 135) = (135-145)/23
= -10/23 = -0.4348
= P ( Z <-0.4348) From Standard Normal Table
= 0.33186
P(X < 155) = (155-145)/23
= 10/23 = 0.4348
= P ( Z <0.4348) From Standard Normal Table
= 0.66814
P(135 < X < 155) = 0.66814-0.33186 = 0.3363                  

b)

To find P(a <= Z <=b) = F(b) - F(a)
P(X < 135) = (135-145)/23/ Sqrt ( 16 )
= -10/5.75
= -1.7391
= P ( Z <-1.7391) From Standard Normal Table
= 0.04101
P(X < 155) = (155-145)/23/ Sqrt ( 16 )
= 10/5.75 = 1.7391
= P ( Z <1.7391) From Standard Normal Table
= 0.95899
P(135 < X < 155) = 0.95899-0.04101 = 0.918                  

d)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 135) = (135-145)/23/ Sqrt ( 36 )
= -10/3.8333
= -2.6087
= P ( Z <-2.6087) From Standard Normal Table
= 0.00454
P(X < 155) = (155-145)/23/ Sqrt ( 36 )
= 10/3.8333 = 2.6087
= P ( Z <2.6087) From Standard Normal Table
= 0.99546
P(135 < X < 155) = 0.99546-0.00454 = 0.9909                  

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