7. The yearly grain consumption of a Wool sheep has a mean of 1780 pounds and st

Question

7. The yearly grain consumption of a Wool sheep has a mean of 1780 pounds and standard deviation 320 pounds.

a) If a sample of 100 sheep is taken, then, the average of the grain consumptions x follows an approximate normal distribution. Find the mean and standard deviation.

(Ans. Mean =1780, Standard Dev = 32.)

b) What is the probability that, from the sample of 100 sheep, the average exceeds x 1828 pounds? (Ans. 6.68%)

8. The NSSL in Oklahoma has determined that the probability a tornado will occur in the state, in a day, is 22%.

Compute the following probabilities.

a) Out of 90 observed days, 25 or more tornados occur in the state.

(Ans. 11.6 %)

b) Out of 90 days observed, fewer than fifteen tornados occur in the state.

(Ans. 9.01%)

*I have the correct answers shown, but would really like a step-by-step of how to get there. Please be as thorough as possible so I can learn from your example. Also what is this called so I can look up the lesson online? Thank you so much!

7. The yearly grain consumption of a wool sheep has a mean = 1780 pounds and standard deviation =320 pounds. If a sample of 100 sheep is taken, then, the average of the grain consumptions x follows an approximate normal distribution. Find the mean and standard deviation Ans. Mean = 1780, Standard Dev 32. What is the probability that, from the sample of 100 sheep, the average x exceeds 1828 pounds? Ans. 6.68% a) b) 8. The NSSL in Oklahoma has determined that the probability a tornado will occur in the state, in a day, is 22%. Compute the following probabilities. a) Out of 90 observed days, 25 or more tornados occur in the state. Ans. 1 1.6 % Out of 90 days observed, fewer than fifteen tornados occur in the state. Ans. 9.01% b)

Explanation / Answer

here you can study central limit theorum for above and use of standard z distribution with table to solve above

here zscore =(X-mean)/std error

a) mean =mean of population =1780

std error of mean =std deviation/(n)1/2 =320/(100)1/2 =32

b) here z score =(1828-1780)/32 =1.5

from normal value table probability for Z>1.5 =0.0668 ~ 6.68%

8)

a)for this problem you need to study normal approximation of binomial distribution

for proportion p=0.22

mean =np =90*0.22 =19.8

and std deviation =3.93

hence P(X>=25) =P(Z>(24.5-19.8)/3.93)=P(Z>0.8841) =0.116 ~11.6% ( from normal value table and 0.5 is reduced for continuity correction)

b) P(X<15) =P(Z<(14.5-19.8)/3.93)=P(Z<-1.3486)=0.09 ~ 9.00% ( from normal value table and 0.5 is reduced for continuity correction)

please revert for further clariifcation.

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