Question

A spaceship has an approximately normal distribution with mean of 521000 spacejugs and a standard deviation of 42000 spacejugs.

a) What is the probability it takes more than 605000 spacejugs of fuel to launch a spaceship?

b) What is the probability it takes between 450,000 and 500000 spacejugs to launch a spaceship?

c) Find the 14th percentile (the point corresponding to the lowest 14%) of fuel used to launch spaceships

Answer 1

a] Solution :

Given ,

mean = = 521000

standard deviation = = 42000

P(x >605000 ) = 1 - P(x<605000 )

= 1 - P[(x -) / < (605000-521000) /42000 ]

= 1 - P(z <2 )

Using z table

= 1 - 0.9772

= 0.0228

probability= 0.0228   

b]

P(450000< x < 500000) = P[(450000-521000) /42000 < (x - ) / < (500000-521000) /42000 )]

= P(-1.69 < Z <-0.5 )

= P(Z < -0.5) - P(Z <-1.69 )

Using z table   

= 0.3085-0.0455

   probability= 0.263

c]

Using standard normal table,

P(Z < z) = 14%

=(Z < z) = 0.14  

= P(Z < z ) = 0.14

z =-1.08   

Using z-score formula  

x = z +

x = -1.08*42000+521000

x = 475640