I need help with my statistics homework! Can anyone help? Not just answers, please explain!?

1. There is a 0.39 prob. that John will buy a car, a 0.73 prob. that Mary will buy a car, and a 0.36 prob. that both will buy a car. Find the prob. that neither will buy a car.





2. Find the prob. using the std. normal distribution.


P(1.12%26lt;z%26lt;1.43)





3. During October, the avg temp of Whitman Lake is 53.2 and the std. dev. is 2.3. Assume the variable is normally distributed. For a randomly selected day in October, find the prob. that the temperature will be as follows:


a. Above 54


b. Below 60


c. Between 49 and 55


d. If the lake temperature were above 60, would you call it very warm?





4. To qualify for officers' training, recruits are tested for stress tolerance. Scores are normally distributed, with a mean of 62 and a std. dev. of 8. If only the top 15% of recruits are selected, find the cutoff score.





5. The avg public school has 458 kids. Assume the std dev is 97. If a random sample of 36 schools is selected, find the prob. that the # of kids enrolled is between 450 %26amp; 465.

I need help with my statistics homework! Can anyone help? Not just answers, please explain!?
1.P(John) + P(Mary) - P(John)P(Mary) = 0.39 + 0.73 - (0.39)(0.73) = 0.8353 is the probability that at least one of them will buy a car. So the probability that neither will by a car is 1 - 0.8353 = 0.1647


2. P(z %26gt; 1.12) = 0.8686; P(z %26lt; 1.43) =0.9236, so


P(1.12 %26lt; z %26lt; 1.43) = 0.9236 - 0.8686 = 0.0550


3.a. P(x %26gt; 54) = 1 - P(x %26lt; 54) Calculate the z-value for P(x %26lt; 54):


Z = (x - µ) /σ = (54 - 53.2)/2.3 = 0.3478


The probability associated with z = 0.3478 is 0.636


So P(x %26gt; 54) = 1 - 0.636 = 0.364


3.b The z-value associated with P(x %26lt; 60) is Z = (60 - 53.2)/2.3


=2.96 and from the standard normal table P(x %26lt; 60) = 0.998


c. You do this using 2 and 3a and 3b as models.


d. The P(x %26gt; 60) = 1 - P(x %26lt; 60) = 1 - 0.998 = 0.002 This is a ver small probability, so you can conclude that the lake is exceptionlly warm.


4. For this problem you have to find the z-value in the table associated with 0.85 (probability of not qualifying) and then solve for x using the equation: Z = (x -µ)/σ


5. We don't know if the distribution of enrolled kids is normally distributed, so the central limit theorem will help.


For this problem you use the equation Z = (x - µ)/(σ/√n).


Find the two z-values 450 and 465 for x and you're given µ = 458 and σ = 97. Then find the difference between the probabilities associated with the z-values. Solve similar to 3c.