Question

The number of accidents per week at a hazardous intersection varies with mean 2.0 and standard deviation 1.2. This distribution takes only whole-number values, so it is certainly not normal. (a) Let x be the mean number of accidents per week at the intersection during a year (52 weeks). What is the approximate distribution of x according to the Central Limit Theorem? μx = σx = (b) What is the approximate probability that x is less than 2? (c) What is the approximate probability that there are fewer than 120 accidents at the intersection in a year? (Hint: Restate this event in terms of x.)

Answer 1

Answer:

a.

μ= 2.0

σ/√n= 1.664

b.

P(X < 2) = 0.05

c.

P(X < 120)= 1

Step-by-step explanation:

Hello!

The study variable is:

X: Number of accidents that occur in an intersection during a period of 52 weeks.

This variable has an unknown distribution but it is known that is mean is μ= 2.0 and its standard deviation is σ= 1.2.

The central limit theorem states that if your sample is big enough (n ≥ 30)even if the distribution of the study variable is unknown, you can aproximate the distribution of the sample mean to normal, symbolically:

X[bar]≈N(μ; σ²/n)

The mean is the same value as the variable

μ= 2.0

And it's variance

σ²/n= (1.2)²/52= 2.769

Standard deviation

σ/√n= √(σ²/n)= 1.664

b.

P(X < 2) = P(Z < [(2 -2)/1.66])= P(Z< 0) = 0.5

c.

P(X < 120) = P(Z < [(120-2)/1.66]) = P(Z < 71.08) ≅ 1

I hope it helps!