4. Given a normal distribution with mean= 100 and standard deviation-20, if a sa

Question

4. Given a normal distribution with mean= 100 and standard deviation-20, if a sample of n = 25 is selected a. What is the probability that x-bar is less than 952 b. What is the probability that x-bar is between 95 and 97.5? c. What is the probability that x-bar is above 102.2? d. What is the probability that x-bar is between 99 and 101? Given a normal distribution with mean-50 and standard deviation- 5, if a sample of n 49 is selected. a. What is the probability that x-bar is less than 47? b. What is the probability that x-bar is between 47 and 49.5? c. What is the probability that x-bar is above 51.1? d. What is the probability that x-bar is between 49 and 51? 5. 6. The following data represent the number of days absent per year in a population of six employees of a small company a. Assuming that you sample without replacement, select all possible samples of size 2 and set up the sampling distribution of the mean. Compute the mean of all the sample means and also compute the population mean. Are they equal? What is this property called? b. Conduct (a) for all possible samples of size 3 Plastic bags used for packaging produce are manufactured so that the breaking strength of the bag is normally distributed with a mean of 5 pounds per square inch and a standard deviation of 1.5 pounds per square inch 7. a. If a sample of 25 bags is selected, what is the probability that the average breaking strength is between 5 and 5.5 pounds per square inch? b. If a sample of 25 bags is selected, what is the probability that the average breaking strength is less than 4.6 pounds per square inch? c. If a sample of 25 bags is selected, what is the probability that the average breaking strength is between 4.2 and 4.5 pounds per square inch?

Explanation / Answer

1. Solition:-
a. P(X > 1.58) = 0.0571

b. P(X < 1.85) = 0.9678

c. P( 1.58 < X < 1.85) = 0.0249

d. 1 - P( 1.58 < x < 1.85) = 0.9751

e. P(-1.58 < and 1.85) = 0.9107

f. 1 - P(-1.58 < and 1.85) = 0.089

2 solution:-

given that mean = 5 sd = 1.5

a. P(5 < X < 5.5) = P( (5-5)/1.5 < Z < (5.5-5)/1.5 )
= P( 0 < Z < 0.3333)
= 0.1293

b. P(3.2 < X 4.2) = P( (3.2-5)/1.5 < Z < (4.2-5)/1.5 )
= P( -1.2 < Z < -0.5333 )
= 0.183

c. P(X 3.6) = P( Z (3.6-5)/1.5 )
= P( Z -0.9333 )
= 0.8238

d. P(X < 3.17) = P(Z < (3.17-5)/1.5 )
= p( Z < -1.22)
= 0.1112

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