Probability problems?

When it rains, there's a chance my newspaper will be wet when I pick it up off my driveway in the morning, even though it's placed in a plastic bag. Experience shows that the probability of a wet paper on a rainy day is 0.26. Out of 50 rainy days in a year, let Y = the number of times my paper is wet. Assume Y is binomial. Use the normal approximation to estimate the probability that the paper is wet more than 18 days , i.e., P(Y %26gt; 18). Give your answer to 3 decimal places





Theresa bought a packet of seeds so she can grow a flower garden. She will start them indoors and plant them outside when the weather is warmer. The packet contains 64 seeds, and Theresa knows from experience that 90% of them will sprout. She plants them all. Assuming that seeds are independent, the number that sprout will be a binomial random variable with mean= ? and standard deviation= ?





any help with either would be great.

Probability problems?
Let Xb be the days the paper is wet. Xb has the binomial distribution with n = 50 trials and success probability p = 0.26



In general, if X has the binomial distribution with n trials and a success probability of p then

P[Xb = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)

for values of x = 0, 1, 2, ..., n

P[Xb = x] = 0 for any other value of x.



To use the normal approximation to the binomial you must first validate that you have more than 10 expected successes and 10 expected failures. In other words, you need to have n * p %26gt; 10 and n * (1-p) %26gt; 10.



Some authors will say you only need 5 expected successes and 5 expected failures to use this approximation. If you are working towards the center of the distribution then this condition should be sufficient. However, the approximations in the tails of the distribution will be weaker especially if the success probability is low or high. Using 10 expected successes and 10 expected failures is a more conservative approach but will allow for better approximations especially when p is small or p is large.



In this case you have:

n * p = 50 * 0.26 = 13 expected success

n * (1 - p) = 50 * 0.74 = 37 expected failures



We have checked and confirmed that there are enough expected successes and expected failures. Now we can move on to the rest of the work.



If Xb ~ Binomial(n, p) then we can approximate probabilities using the normal distribution where Xn is normal with mean μ = n * p, variance σ2 = n * p * (1-p), and standard deviation σ



Xb ~ Binomial(n = 50 , p = 0.26 )

Xn ~ Normal( μ = 13 , σ2 = 9.62 )

Xn ~ Normal( μ = 13 , σ = 3.101612 )



I have noted two different notations for the Normal distribution, one using the variance and one using the standard deviation. In most textbooks and in most of the literature, the parameters used to denote the Normal distribution are the mean and the variance. In most software programs, the standard notation is to use the mean and the standard deviation.



The probabilities are approximated using a continuity correction. We need to use a continuity correction because we are estimating discrete probabilities with a continuous distribution. The best way to make sure you use the correct continuity correction is to draw out a small histogram of the binomial distribution and shade in the values you need. The continuity correction accounts for the area of the boxes that would be missing or would be extra under the normal curve.



P( Xb %26lt; x) ≈ P( Xn %26lt; (x - 0.5) )

P( Xb %26gt; x) ≈ P( Xn %26gt; (x + 0.5) )

P( Xb ≤ x) ≈ P( Xn ≤ (x + 0.5) )

P( Xb ≥ x) ≈ P( Xn ≥ (x - 0.5) )

P( Xb = x) ≈ P( (x - 0.5) %26lt; Xn %26lt; (x + 0.5) )

P( a ≤ Xb ≤ b ) ≈ P( (a - 0.5) %26lt; Xn %26lt; (b + 0.5) )

P( a ≤ Xb %26lt; b ) ≈ P( (a - 0.5) %26lt; Xn %26lt; (b - 0.5) )

P( a %26lt; Xb ≤ b ) ≈ P( (a + 0.5) %26lt; Xn %26lt; (b + 0.5) )

P( a %26lt; Xb %26lt; b ) ≈ P( (a + 0.5) %26lt; Xn %26lt; (b - 0.5) )



In the work that follows Xb has the binomial distribution, Xn has the normal distribution and Z has the standard normal distribution.



Remember that for any normal random variable Xn, you can transform it into standard units via: Z = (Xn - μ ) / σ





P( Xb %26gt; 18 ) =



50

∑ P(Xb = x) = 0.04183087

x = 19



≈ P( Xn %26gt; 18.5 )

= P( Z %26gt; ( 18.5 - 13 ) / 3.101612 )

= P( Z %26gt; 1.773271 )

= 0.03809189



== -- == -- == -- == -- ==



Let X be the number of seeds that will germinate. X has the binomial distribution with n = 64 trials and success probability p = 0.9



In general, if X has the binomial distribution with n trials and a success probability of p then

P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)

for values of x = 0, 1, 2, ..., n

P[X = x] = 0 for any other value of x.



The probability mass function is derived by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.

Or, in other words, the binomial is the sum of n independent and identically distributed Bernoulli trials.



X ~ Binomial( n , p )



the mean of the binomial distribution is n * p = 57.6

the variance of the binomial distribution is n * p * (1 - p) = 5.76

the standard deviation is the square root of the variance = √ ( n * p * (1 - p)) = 2.4
Reply:if you have a graphing calculator, the first problem is very easy. On a TI-83, you press 2nd--%26gt;DISTR then go to binomcdf and enter your given information, so you would enter binomcdf(0.26,18), then subtract your answer from 1 because you want the probability that you get MORE than 18 days.



For the second problem, the mean of a binomial variable is written as E(x) or the Expected Value. To calculate the expected value, you multiply the number of trials (in this case, you have 64 seeds, meaning 64 trials) by the probability of success (0.9) The answer is 57.6. The standard deviation of a binomial variable is square root of np(1-p) where again n is the number of trials and p is the probability of success. in this problem you would take the square root of (64)(.9)(.1)