Question

(3 points) Blades of grass Suppose that the heights of blades of grass are Normally distributed and independent, with each height having expected value 4 inches and standard deviation
0.75
inches. A child picks the blades of grass. (a) ( 1 point) What is the probability that the height of 1 blade of grass is 5 inches or taller? (b) (1 point) How many blades of grass does the child expect to pick until the first blade of grass appears with height 5 inches or taller? (c) (1 point) Now, instead of the setup in
1 b
, suppose that the child picks exactly 10 blades of grass. What is the expected number of blades of grass that have height 5 inches or taller, within this collection of 10 blades of grass?

Answer 1

The final answer is:

a) P( Y < 42.5 )  = 0.8541

b) P( 39.5 < Y < 40.5 ) = 0.1670.

What is the normal distribution?

A continuous probability distribution for a real-valued random variable in statistics is known as a normal distribution or Gaussian distribution.

If x follows a normal distribution with mean μ and standard deviation σ then the distribution of

$\sum_{i =1}^{n}x_{i}$  follows an approximately normal distribution with a mean $n\mu$ and standard deviation $\sqrt{n }\sigma$

let x be the height of blades of grass

x follows normal distribution with mean = μ = 4 and standard deviation = σ = 0.75.

Y = x1 + x2 +...........+x10

$Y = \sum_{i =1}^{10}x_{i}$

Distribution of Y is normal with,

Mean = $\mu _{y}=10*4 = 40$ and standard deviation $= \sigma _{y}=\sqrt{10}*0.75 = 2.3717$

a)

P( Y < 42.5 )

Using normal distribution formmula,

$f(x)= {\frac{1}{\sigma\sqrt{2\pi}}}e^{- {\frac {1}{2}} (\frac {x-\mu}{\sigma})^2}$

=NORMDIST( x, mean, SD , 1 )      

=NORMDIST(42.5, 40, 2.3717, 1 )

=0.8541

P( Y < 42.5 )  = 0.8541

b)

P( 39.5 < Y < 40.5 ) = P( Y < 40.5 ) - P( Y < 39.5 )

Using normal distribution formmula,

$f(x)= {\frac{1}{\sigma\sqrt{2\pi}}}e^{- {\frac {1}{2}} (\frac {x-\mu}{\sigma})^2}$

P( Y < 40.5 )  =NORMDIST(40.5, 40, 2.3717, 1 ) = 0.5835

P( Y < 39.5 ) = NORMDIST(39.5, 40, 2.3717, 1 ) = 0.4165

P( 39.5 < Y < 40.5 ) = 0.5835 - 0.4165  = 0.1670

P( 39.5 < Y < 40.5 ) = 0.1670

Hence, the final answer is:

a) P( Y < 42.5 )  = 0.8541

b) P( 39.5 < Y < 40.5 ) = 0.1670.

To learn more about the normal distribution visit,

brainly.com/question/4079902

#SPJ4