Answer:
a) The probability that a random movie is between 1.8 and 2.0 hours = 0.2586.
b) The probability that a random movie is longer than 2.3 hours is 0.0918.
c) The length of movie that is shorter than 94% of the movies is 1.4 hours
Explanation:
In the above question, we would solve it using z score formula
z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation
a) A random movie is between 1.8 and 2.0 hours
z = (x-μ)/σ,
x1 = 1.8,
x2 = 2.0
μ is the population mean = 1.9
σ is the population standard deviation = 0.3
z1 = (1.8 - 1.9)/0.3
z1 = -1/0.3
z1 = -0.33333
Using the z score table
P(z1 = -0.33) = 0.3707
z2 = (2.0 - 1.9)/0.3
z1 = 1/0.3
z1 = 0.33333
p(z2 = 0.33) = 0.6293
= P(- 0.33 ≤ z ≤ 0.33)
= 0.6293 - 0.3707
= 0.2586
The probability that a random movie is between 1.8 and 2.0 hours = 0.2586
b) A movie is longer than 2.3 hours
z = (x-μ)/σ,
x1 = 2.3
μ is the population mean = 1.9
σ is the population standard deviation = 0.3
z = (2.3 - 1.9)/0.3
z = 4/0.3
z = 1.33333
P(z = 1.33) = 0.90824
P(x>2.3) = = 1 - 0.90824
= 0.091759
≈ 0.0918
The probability that a random movie is longer than 2.3 hours is 0.0918.
3) The length of movie that is shorter than 94% of the movies.
z = (x-μ)/σ
Probability (z ) = 94% = 0.94
Movie that is shorter than 0.94
= P(1 - 0.94) = P(0.06)
Finding the P (x< 0.06) = -1.555
≈ -1.56
μ is the population mean = 1.9
σ is the population standard deviation = 0.3
-1.56 = (x - 1.9)/ 0.3
Cross multiply
-1.56 × 0.3 = x - 1.9
- 0.468 + 1.9 = x
= 1.432 hours
≈ 1.4 hours
Therefore, the length of movie that is shorter than 94% of the movies is 1.4 hours