Question

A road averages 2,895 vehicles per day with a standard deviation of 615 vehicles per day. A traffic counter was used on this road on 34 days that were randomly selected.

a. What is the probability that the sample mean is less than 2,700 vehicles per​ day?

b. What is the probability that the sample mean is more than 2,900 vehicles per​ day?

c. What is the probability that the sample mean is between 2,800 and 3,000 vehicles per​ day?

d. Suppose the sample mean was 3,100 vehicles per day. Does this result support the stated population mean for this​ road?

The probability that the average vehicles per day is more than 3,100 is _____.  The result (does or does not)

support the findings because this probability is (less than or equal to; greater than) 0.05.

​(Type an integer or decimal rounded to four decimal places as​ needed.)

Answer 1

a)
Here, μ = 2895, σ = 105.4716 and x = 2700. We need to compute P(X <= 2700). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (2700 - 2895)/105.4716 = -1.85

Therefore,
P(X <= 2700) = P(z <= (2700 - 2895)/105.4716)
= P(z <= -1.85)
= 0.0322

b)
Here, μ = 2895, σ = 105.4716 and x = 2900. We need to compute P(X >= 2900). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (2900 - 2895)/105.4716 = 0.05

Therefore,
P(X >= 2900) = P(z <= (2900 - 2895)/105.4716)
= P(z >= 0.05)
= 1 - 0.5199 = 0.4801

c)

Here, μ = 2895, σ = 105.4716, x1 = 2800 and x2 = 3000. We need to compute P(2800<= X <= 3000). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (2800 - 2895)/105.4716 = -0.9
z2 = (3000 - 2895)/105.4716 = 1

Therefore, we get
P(2800 <= X <= 3000) = P((3000 - 2895)/105.4716) <= z <= (3000 - 2895)/105.4716)
= P(-0.9 <= z <= 1) = P(z <= 1) - P(z <= -0.9)
= 0.8413 - 0.1841
= 0.6572

d)

Here, μ = 2895, σ = 105.4716 and x = 3100. We need to compute P(X >= 3100). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (3100 - 2895)/105.4716 = 1.94

Therefore,
P(X >= 3100) = P(z <= (3100 - 2895)/105.4716)
= P(z >= 1.94)
= 1 - 0.9738 = 0.0262

The probability that the average vehicles per day is more than 3,100 is0.0262. The result (does

support the findings because this probability is (less than or equal to 0.05.