7. Suppose that the mean weight of my books is 1 kg and the standard deviation i

Question

7. Suppose that the mean weight of my books is 1 kg and the standard deviation is 0.4 kg. I am moving to another state, so I pack my books in boxes, 64 books in each box. Find the probability that a randomly selected box will weigh more than 57.6 kg? (Hint: Use the Central Limit Theorem) (a) 0.0228 (b) 0.4013 (c) 0.5987 (d) 0.9772 (e) None of these

8. A farmer divided his field into a large number of small plots of the same size. The amount of yield per plot has a mean of 100 bushels with a standard deviation of 8 bushels. If 64 plots are selected at random, find the probability that the mean yield will be between 98 and 102 bushels. (a) 0.4013 (b) 0.5987 (c) 0.9544 (d) 0.9772 (e) None of these

7. Suppose that the mean weight of my books is 1 kg and the standard deviation is 0.4 kg. I am moving to another state, so I pack my books 64 books in each box. Find the probability that a randomly selected box will weigh more than 57.6 kg? (Hint: Use the Central Limit Theorem) (c) 0.5987 (a) 0.0228 (b) 0.4013 (d) 0.9772 (e) None of these 8. A farmer divided his field into a large number of small plots of the same size. The amount of yield per plot has a mean of 100 bushels with a standard deviation of 8 bushels. If 64 plots are selected at random, find the probability that the mean yield will be between 98 and 102 bushels (c) 0.9544 (a) 0.4013 (b) 0.5987 (d) 0.9772 (e) None of these

Explanation / Answer

7.

weights of the books are normally distributed with mean 1kg and standard deviation 0.4 kg.

when we packed the books in a box, then the weights of the box will be normally distributed random variable with following properties......

The mean of the distribution will be sum of the mean of all the books contained in the box= 64*1=64 kg

Variance of the distribution will be sum of the variance of all the books contained in the box =64*0.42

So the standard deviation of the box will be sqrt(64*0.42) = 8 * 0.4=3.2 kg

let W is the random variables denotes the weight of the box. so W~N(64,3.22)

Now we have to find p(W>57.6)= p(Z>(57.6-64)/3.2) = p(Z> -2)

=p(Z<2) [from the symetric nature of normal distribution)

=0.9772 [from the Z table]

so the correct answer is d

8.

Here we take a sample of 64 plots. So the sampling distribution of the sample mean will have a normal distribution

with mean 100 and standard deviation 8/sqrt(64) = 8/8 =1 [According to Central Limits theorem]

if X is the random variable, then we have to find....

P(98< X < 102) = P( (98-100)/1 < Z < (102-100)/1 )= P( -2 < Z <2)= P(Z < 2) - P(Z<-2)

=0.97725 - 0.02275 = 0.95450

So the correct answer is c

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