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Neko [114]
3 years ago
12

Assume the speed of vehicles along a stretch of I-10 has an approximately normal distribution with a mean of 81 mph and a standa

rd 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?

Mathematics
1 answer:
Pachacha [2.7K]3 years ago
3 0

Answer:

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!

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GarryVolchara [31]

Answer:

\mu_p -\sigma_p = 0.74-0.0219=0.718

\mu_p +\sigma_p = 0.74+0.0219=0.762

68% of the rates are expected to be betwen 0.718 and 0.762

\mu_p -2*\sigma_p = 0.74-2*0.0219=0.696

\mu_p +2*\sigma_p = 0.74+2*0.0219=0.784

95% of the rates are expected to be betwen 0.696 and 0.784

\mu_p -3*\sigma_p = 0.74-3*0.0219=0.674

\mu_p +3*\sigma_p = 0.74+3*0.0219=0.806

99.7% of the rates are expected to be betwen 0.674 and 0.806

Step-by-step explanation:

Check for conditions

For this case in order to use the normal distribution for this case or the 68-95-99.7% rule we need to satisfy 3 conditions:

a) Independence : we assume that the random sample of 400 students each student is independent from the other.

b) 10% condition: We assume that the sample size on this case 400 is less than 10% of the real population size.

c) np= 400*0.74= 296>10

n(1-p) = 400*(1-0.74)=104>10

So then we have all the conditions satisfied.

Solution to the problem

For this case we know that the distribution for the population proportion is given by:

p \sim N(p, \sqrt{\frac{p(1-p)}{n}})

So then:

\mu_p = 0.74

\sigma_p =\sqrt{\frac{0.74(1-0.74)}{400}}=0.0219

The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).

\mu_p -\sigma_p = 0.74-0.0219=0.718

\mu_p +\sigma_p = 0.74+0.0219=0.762

68% of the rates are expected to be betwen 0.718 and 0.762

\mu_p -2*\sigma_p = 0.74-2*0.0219=0.696

\mu_p +2*\sigma_p = 0.74+2*0.0219=0.784

95% of the rates are expected to be betwen 0.696 and 0.784

\mu_p -3*\sigma_p = 0.74-3*0.0219=0.674

\mu_p +3*\sigma_p = 0.74+3*0.0219=0.806

99.7% of the rates are expected to be betwen 0.674 and 0.806

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