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the Mean L hows the e) Fiad the sample standard devistion 내ちょ 225 ndopted anad live 7 oe moce yean (b) The probability of a plam ping towed 38%,sad the probabity ofbeaal0% was-th 34% Th. prohallity of that sa-typedpla plant will go to seed ar survive the wister? AbMketball player makm 70% ofthe bee throws beshoot. Seppose th be trm is bmthns (a) What in the probability that be will maks moee thsa shwe (b) Find the expected value and the standard deviation Let z be a randen variable that represeats the length d tae a uus·stala e vie。 reeponse paper. It was found that z has an apprcximabely normal distribation with 72 hous and standard deviation σ-1.8 hour, (a) What is the probability that it takes at least 5 hours for a stadent to write areponse peper (b) Suppose 20 atudents are selected st random. What is the probability that the mess tie of writing a paper for these 20 students is not more than 8 hours? Collette is self-employed, selling cosmetics at home parties. Ste wasts to etimate the sversge amount a client spends per year at these parties. A random sample of 16 receipta lad a mes of -$340 with a standard deviation ofs-$60.15. Pind a 90% confdence imberval tr-seas ano=-pet by all clients
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Answer #1

Dear Student,

Please post the question one at a time.

4) X is normally distributed with population mean := μ=7.2

and Standard deviation = σ = 1.8

a) The probability that it takes at least 5 hours to write a response paper can be represented as = P(X 2 5)

we will convert this to standard Z score to find the probability

Z=rac{X-mu}{sigma}

5-7 1.8

P(Z < 1.11) = 0.8665

P(X > 5) = 1-P(Z < 1.11) = 1-0.8665= 0.13335

b) If 20 Students are selected at random then by definition of centra limit theorem,

the mean time of the students is normally distributed with

population mean := μ=7.2

and Standard deviation = σ = 1.8

now we will convert this to standard Z score.

Vn 7t

Now we have to calculate the probability that the sample mean is not more than 8

8-7 P(X < 8)-P(Z < T8 ) P(Z < 2.48) 0.9934 V20

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