Police use a radar unit is used to measure speeds of cars on a freeway. The speeds are normally distributed with a mean of 90 km/hr and a standard deviation of 10 km/hr. (a) What is the probability that a car picked at random is traveling at more than 100 km/hr? (b) Since normal distributions technically have a domain of (−[infinity],[infinity]), this distribution allows the speed of a car to be negative. But does it really!? Determine (or try to) the probability that the speed of a car is negative and decide for yourself if this is at all possible.

Answer:A. P(x≥100)=0.1587B. P(x≤0)≈0Step-by-step explanation:A. Cause we know the distribution of the data, the method used to solve it is called "Normalization" and we need to have the Mean and the Standard deviation of the data. The method consist in the following equationP(x≤a)=P( z=((x-μ)/σ) ≤ b=((a-μ)/σ) )Considering μ as the Mean and σ as the Standard deviation. At first, we had a probability in the normal distribution with Mean=90 and STD=10 but that kind of exercises is not meant to find that probability directly but by using this process. After we normalize the probability, now we have a probability in a specific normal distribution that has Mean=0 and STD=1 and the difference with what we had before is that now we are able to use tools to find probabilities in a normal standard distribution. My favorite of them is a chart that show the approximate values of a lot of probabilities (i attached it to this answer). I´m going to explain point A as an example:We look for the probability that P(x≥100), but we don´t have an easy method to use there, so we normalize:P(x≥100)=P( (x-μ)/σ ≥ (100-μ)/σ )P(x≥100)=P( z ≥ (100-90)/10 )P(x≥100)=P( z ≥ 1 )And now we are able to use the chart, let me explain: First, the chart only works with P(z ≤ b), so we have to change it with properties of probabilities before using the table.P(z≥1)=1-P(z≤1)And finally we use the chart: the value of P(z≤1) is in the table, we look for the row with +1 and the column with the decimal part (in this case 0) and with coordinates (1,0) there´s the value:P(z≤1)=0.8413But we need P(z≥1) so we use the previous equalityP(z≥1)=1-P(z≤1)P(z≥1)=1-0.8413P(z≥1)=0.1587Because P(x≥100)=P(z≥1), our final answer is 0.1587B. We use the same process to try to understand what the probability of P(x≤0) represents.P(x≤0)=P(z≤ (0-90)/10)P(x≤0)=P( z ≤ -9 )But when we try to look for its value in the chart It isn´t even there, what could it mean?A normal distribution function is always increasing, that means that "a≤b if and only if P(x≤a) ≤ P(x≤b)". so we conclude:P(z≤-9) ≤ P(z≤-3) (The lowest probability in the chart)P(z≤-9) ≤ 0.0013P(z≤-9) is way lower than 0.0013 (they aren´t even close) but we know that probability is always positive,  and because of that:P(x≤0)=P(z≤-9)≈0