Assume the speed of vehicles along a stretch of I-10 has an approximately normal distribution with a mean of 81 mph and a standard deviation of 8 mph. The speed limit is 65. If you pick a car on the highway at random, what is the probability the vehicles is going less than or equal to the speed limit?

Step-by-step explanation:

Hello!

The variable of interest is

X: speed of a vehicle along a stretch of I-10 (mph)

This variable has a normal distribution with mean μ= 81 mph and a standard deviation σ= 8 mph.

The speed limit in the said stretch is 65 mph.

You need to calculate the probability of picking a car at random and its speed be at most 65 mph, symbolically:

P(X≤65)

To reach the probability, you need to use the standard normal distribution. To standardize the value fo X you have to subtract the value of μ and then divide it by σ:

P(Z≤(65-81)/8)= P(Z≤-2.00)

Now you look for the corresponding probability in the table of the standard normal distribution, since the value is negative you have to use the left entry. The integer and first decimal numbers are in the first column and the second decimal number is in the first row.

P(Z≤-2.00)= 0.0228

I hope it helps!

x = 12

y = 15

step-by-step explanation:

the answer is ! ?