Question

The average monthly cable bill in 2016 has been reported to be ​$92. Assume monthly cable bills follow a normal distribution with a standard deviation of ​$9.50.

a. What is the probability that a randomly selected bill will be

1. less than​ $90​?

2. less than ​$100​?

3. exactly ​$100​?

4. between ​$85 and ​$95​?

Which monthly cable bill represents the 55th percentile?

Answer 1

1)

Here, μ = 92, σ = 9.5 and x = 90. We need to compute P(X <= 90). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (90 - 92)/9.5 = -0.21

Therefore,
P(X <= 90) = P(z <= (90 - 92)/9.5)
= P(z <= -0.21)
= 0.4168

2)

Here, μ = 92, σ = 9.5 and x = 100. We need to compute P(X <= 100). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (100 - 92)/9.5 = 0.84

Therefore,
P(X <= 100) = P(z <= (100 - 92)/9.5)
= P(z <= 0.84)
= 0.7995

3)

Here, μ = 92, σ = 9.5, x1 = 99.5 and x2 = 100.5. We need to compute P(99.5<= X <= 100.5). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (99.5 - 92)/9.5 = 0.79
z2 = (100.5 - 92)/9.5 = 0.89

Therefore, we get
P(99.5 <= X <= 100.5) = P((100.5 - 92)/9.5) <= z <= (100.5 - 92)/9.5)
= P(0.79 <= z <= 0.89) = P(z <= 0.89) - P(z <= 0.79)
= 0.8133 - 0.7852
= 0.0281

4)

Here, μ = 92, σ = 9.5, x1 = 85 and x2 = 95. We need to compute P(85<= X <= 95). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (85 - 92)/9.5 = -0.74
z2 = (95 - 92)/9.5 = 0.32

Therefore, we get
P(85 <= X <= 95) = P((95 - 92)/9.5) <= z <= (95 - 92)/9.5)
= P(-0.74 <= z <= 0.32) = P(z <= 0.32) - P(z <= -0.74)
= 0.6255 - 0.2296
= 0.3959

5)

z value t 55% = 0.13
z = (x - mean)/s
0.13 = (x - 92)/9.5
x = 9.5 * 0.13 + 92
x = 93.24