Question

nx P(x) = ckp (-p)" np = E ) (with improved notulen k additional formulas). 22. in a given your dev be $100 e r, the average annuelley of NFL f all player was $189.000 tandard $20.500. Ir sample of 50 players was taken the probability that the sample mean will be 5192,000 or less is: 0.1515. 23. Z- erede sales per customer at home movement store during the past year is 575 with andard deviation of 5 6. The pr i t he sales customer from a sample of 36 customers, taken at random from this population exceda 572 a. 0.9013 b. 0.5000 0.9332 d. 0.0669 e. 09987. 2 24. Rul A retailer finds that the demand for a very popular bound me everages 100 per week with a standard deviation of 20. If the seller wishes to have adeguate stock 975 % of the time, how many of the games must she keep on hand? a. 139.2 b. 100.0 c. 132.9 d. 195.0 c. 125.6 25 A salesman who uses his car extensively finds that his gasoline bills average $125.32 per month with a standard deviation of $49.51. The probability that his bill will be more than $50 a month less than $150 for a single month is: a 0.4357 b. 0.1915. c. 0,6272 d. 0.3728 e. 0.6915. 26 In a continuous probability distribution, the probability that will take on an exact value a. is equal to the height of the curve at that value. b. is calculated using the probability density. c. is always greater than 0. d. is always equal to 0. e. None of these is correct. 27 -1.38 is: Using the standard normal table, the total area between z = 0.00 and 2 a. 0.0494 b. 0.4656 c. 0.1554 d. 0.4162 e. 0.1005 28. The Z-score representing the 10th percentile of the standard normal distribution is: a. 0.67. b. -0.67. c. 1.28 d. -1.28. e. 1.645.

Answer 1

22) Let X be a random variable denoting the average salary of the football players. Let us assume X follows normal distribution. Given, X~N(189000,20500²) .

A sample of size 50 is taken. Thus, the sample mean $\bar{X}$ ~N(189000, 20500²/50)

Therefore, the probability that the sample mean will be $192000 Or less is=

P($\bar{X}$<192000)

=P[{($\bar{X}$-189000)*√50/20500}<{(192000-189000) *√50/20500}]

=P(Z<1.03) [where Z is the standard normal variable]

=0.8485 [from the standard normal table]

Thus, the required probability is=0.8485.

Therefore, the correct option is=(b) .