The random variable, X, representing the number of cherries in a cherry cake, has the following probability distribution: |4 P(X=x) 0.2 |5 7 0.1 0.4 0.3 a) Find the mean u and the variance a of X. b) Find the mean ug and the variance of of the sampling distribution & for random samples of 36 cherry cakes.
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- We are examining farmer production in marginal land. A sample of 100 farmers produce this crop with mean of 100 bushels per acre. Assume a normal distribution. Q1: What is the probability an individual farmer in this population produces 103 or less bushels of corn?Q2: What is the probability the mean bushels of corn produced is 103 or less bushels of corn?Dandelions have effects on crop production and lawn growth and are studied. In one region, the mean number of dandelions per square meter was found to be 10.8.Find the probability of no dandelions in an area of 1 m².P(X=0)=P(X=0)= Find the probability of at least one dandelion in an area of 1 m².P(at least one) = Find the probability of at most two dandelions in an area of 1 m².P(X≤2)=P(X≤2)=As the sample size increases, the mean of the sampling distribution ▼ stays the same. gets larger. gets smaller.
- Assume that the average score on a statistical test is 8.5 and the variance is 2. The test was on the lighter side and the score distribution was very skewed (left skewed). What probability distribution is it reasonable to assume that the average grade of 3 randomly selected students who take the exam follows? 1. Normal distribution with mean 8.5 and variance 2. 2. Normal distribution with an average of 8.5 and a variance of 2/3. 3. Poisson distribution with an average of 8.5 and a distribution of 8. 4. The probability distribution cannot be predicted from the information provided.Thirty percent of the students in a class of 100 are planning to go to graduate school. The Variance of this binomial distribution is: A. 16 B. 21 C. 25 D. 36 Cat1. Service time for a customer coming through a checkout counter in a retail store is a random variable with the mean of 10.0 minutes and standard deviation of 4.0 minutes. Suppose that the distribution of service time is fairly close to a normal distribution. Suppose there are two counters in a store, n₁ = 41, customers in the first line and m₂ = 51 customers in the second line. Find the probability that the difference between the mean service time for the shorter line and the mean service time for the longer X₂ one ¹2 is more than 0.4 minutes. Assume that the service times for each customer can be regarded as independent random variables.
- The weights of steers in a herd are distributed normally. The variance is 40,000 and the mean steer weight is 1400lbs. Find the probability that the weight of a randomly selected steer is between 1537lbs and 1679lbs Round your answer to four decimal places.Suppose that the probability that a patient admitted in a hospital is diagnosedwith a certain type of cancer is 0.03. Suppose that on a given day 10 patients areadmitted and X denotes the number of patients diagnosed with this type of cancer.Determine the probability distribution of the random variable X. Find the meanand the variance of XFind the variance of the probability distribution for the histogram shown.
- Dandelions are studied for their effects on crop production and lawn growth. In one region, the mean number of dandelions per square meter was found to be 11.Find the probability of no dandelions in an area of 1 m².P(X=0)=P(X=0)=Find the probability of at least one dandelion in an area of 1 m².P(at least one) =Find the probability of at most two dandelions in an area of 1 m².P(X≤2)=Let Y1,Y2,...Yn be a random sample from a normal distribution where the mean is 2 and the variance is 4. how large must in order that P(1.9≤Y≤2.1)≥0.99?Suppose that among salmon fish that are longer than 80cm, their weights have a mean of 6.4kg and a standard deviation of 0.5kg. The wholesale price per kg for these salmon fish is $9.46. The fishing company supplies freshly caught salmon to a local restaurant. The restaurant just placed an order for 120 salmon fish. If the fishing company randomly selects 120 salmon that are longer than 80cm for the restaurant, find the probability that the total wholesale price of the 120 salmon exceeds $7,350.