Let x 1 , x 2 , . . . , x 100 denote the actual net weights (in pounds) of 100 r

Question

Let x1, x2, . . . , x100 denote the actual net weights (in pounds) of 100 randomly selected bags of fertilizer. Suppose that the weight of a randomly selected bag has a distribution with mean 35 lb and variance 1 lb2. Let x be the sample mean weight (n = 100).

(a) What is the probability that the sample mean is between 34.75 lb and 35.25 lb? (Round your answer to four decimal places.) P(34.75 x 35.25)

Suppose that two graduating seniors, one a marketing major and one an accounting major, are comparing job offers. The accounting major has an offer for $45,000 per year, and the marketing student has an offer for $43,000 per year. Summary information about the distribution of offers follows:

Then calculate the appropriate z scores. Round the answers to two decimal places.   

(so $45,000 is ---------- standard deviation below the mean) *** please i need answer to this part the most*** (45000 is what standard deviation below the mean?)

Accounting: mean = 46,000 standard deviation = 1200 Marketing: mean = 42,500 standard deviation = 1200 accounting 2 score 45.000 46.000 1200

Explanation / Answer

here sample size=n=100 is quite large

hence the distribution of x,the sample mean weight can be approximated by a normal distribution with

mean=E[x]=E[(x1+x2+..+x100)/100]=35*100/100=35 lbs

variance=V[x]=V[(x1+x2+...+x100)/100]=1*100/1002=1/100

so P[34.75<=x<=35.25]=P(34.75-35)/sqrt(1/100)<=(x-35)/sqrt(1/100)<=(35.25-35)/sqrt(1/100)]

=P[-2.5<=Z<=2.5] where Z~N(0,1)

=P[Z<=2.5]-P[Z<=-2.5]=0.9937903-0.006209665=0.9876 [answer]

let $45000 is k standard deviation below the mean

so 46000-k*1200=45000

or, k=(46000-45000)/1200=-(45000-46000)/1200=- z score

now accounting z score is =(45000-46000)/1200=-0.83

hence $45000 is 0.83 standard deviation below the mean [answer]

marketing z score=(43000-42500)/1200=0.42 [answer]

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