The length of time a full length movie runs from opening to credits is normally distributed with a mean of 1.9 hours and standard deviation of 0.3 hours. Calculate the following: A random movie is between 1.8 and 2.0 hours. A movie is longer than 2.3 hours. The length of movie that is shorter than 94% of the movies

Step-by-step explanation:

given data

mean = 1.9 hours

standard deviation = 0.3 hours

solution

we get here first  random movie between 1.8 and 2.0 hours

so here

P(1.8 < z < 2 )

z = (1.8 - 1.9) ÷ 0.3

z = -0.33

and

z = (2.0 - 1.9) ÷ 0.3

z = 0.33

z = 0.6293

so

P(-0.333 < z < 0.333 )

=  0.26

so random movie is between 1.8 and 2.0 hours long is 0.26

and

A movie is longer than 2.3 hours.

P(x > 2.3)

P(  > )

P (z  >  )

P (z  > 1.333  )

= 0.091

so chance a movie is longer than 2.3 hours is 0.091

and

length of movie that is shorter than 94% of the movies is

P(x > a ) = 0.94

P(x <  a ) = 0.06

so

P( <   )

a = 1.43

so length of the movie that is shorter than 94% of the movies about 1.4 hours.

t = 3

step-by-step explanation:

put t = 3 in this expression :

(3²-1)(3²-2)(3² - 3)(3² - 4) = 8×7×6×5 = 56×30 = 1680