Question

volving the Mean s Le L.т, .. : Tho chart below shows the dats on the number of hits oa perod. website per week over sSwok 55- 115 38 185 225 (a) Find the median. 2is5 ()Find the sample standard deviation s @f3sı .. 145.4), (.,5-143,-)n + (ารซ.14%).thto a_ 14qu)". ard L-To) Piad the sample mean 3& SIeS135+11s-143. Iz'e.en'i (a) Anan a >belter has a 58 adoption rate fr parim Of the puppies who are adopted, 90% live to be 7 yeare or older. Whst is the probability that & rasdomly velected pappy in the sbalter will se adopted and live 7 or more years? (b) The probability of a plant going to seed as 34%. The probability of thai sa ae type of plat urviving the winter is 38%, and the probabllity of both s l0%, what Is the probabilty that ared plant will go to seed or survive the winter? s randomly selected alr 3. A basketball player makes 70% of the free throws he shoots. Suppose that he tris is iroo throws. (a) What is the probability that be will mnake more than 7 theow (b) Find the expected value and the standard devistion. Let z be a random variable that, represents the length of time it takes a studest to write a response paper. It was found that z has an approximately normal distribution with mean u 7.2 hours and standard deviation σ = 1.8 hours. (a) What is the probability that it takes at least 5 bours for a student to write s response paper? (b) Suppose 20 students are selected at random. What is the probability that the mean time of writing a paper for these 20 students is not more than 8 hours? Collette is self-employed, selling cosmetics at home parties. She wants to estimate the sverage amount a client spends per year at these parties. A random sample of 16 receipts had a mean of -$340.70 with a standard deviation of s-860.15. Find a 90% confidence interval for the mean amount spent by all clients.

Answer 1

1. A) Arranging the data in ascending order

38, 115, 155, 185, 225

Median: 155

B) Mean = Sum of the data points/number of points

     Mean = 38+115+155+185+225/5 = 143.6

C) Standard Deviation = ∑(x-mean)²

                                        = (38-143.6)²+(115-143.6)² +(155-143.6)²+(185-143.6)²+(225-143.6)²

                                                                =20.439.14

2. A) Adoption rate = 58% = 0.58

    Probability of those puppies living longer than or equal to 7 years = 0.9

    Probability of puppy being adopted and living more than or equal to 7 years = 0.58 * 0.9 =             0.522

B) Probability of plant going to seed = 0.34

     Probability of a plant surviving winter = 0.38

     Probability of both above conditions happening = 0.1

Probability of Plant to seed or surviving winter (P(A⋃B) = Probability of plant going to seed (P(A)) + Probability of a plant surviving winter (P(B)) - Probability of both above conditions happening (P(A⋂B)

         = 0.34+0.38-0.1

         = 0.62