Question

A) 0 5000 B 4960 0.4000 D .S010 CEO Bear he w as pre d by one or me score of 80 and andar dan 12 Spare me 200 m ix raken on the flowing questions 6) What is the means of the sample who score? A) SS D) 68 E) 92 2 Bascha STA 2200 Sample Final. P 7) What is the standard deviation of the sample mean score! A) 0.62 B) 1,03 C) 0.85 Di 0.97 E) 0.75 8) What is the probability that the sample mean score is more than 787 A) 0.9949 B) 0.9938 C) 0.9881 D) 0.9906 E) 0.9922 9) If the sample size were 100, how would your answer to the previous question change? A) The probability would become larger if the population is normally distributed B) The probability would remain the same C) The probability would become 0.5 D) The probability would become smaller E) The probability would become larger 10) What is the 97 percentile of sample mean score! A) 79.312 B) 80.157 C) 82.258 D) 80.886 E) RI 598 Daily gross sales in those of dollars) were recorded for a same of 20 small businesses in the New York metropolitan area in March 2009 yielding a mear of 283.2 with a sundard devianion of 118.023. Suppose the daily gros sales for all small businesses follow a normal model d e the following questions 11) A 95% confidence interval for the true mean daily gross sales for all small businesses is given by AX (227.96, 338.44) 211.47,314.93) C) (214.58. 331.12)

answer 6-10.. please show all the work..

Answer 1

Answer)

As the population s.d is known we can assume normality

6)

Mean is same as sample mean = 80

7)

S.d = s.d/√n = 12/√200 = 0.85

8)

P(x>78)

Answer)

As the data is normally distributed we can use standard normal z table to estimate the answers

Z = (x-mean)/s.d

Z = (78 - 80)/0.85) = -2.35

From z table, P(z>-2.35) = 0.9906

9)

In case of n = 100

Z = (78-80)/(12/√100) = -1.67

From z table, P(z>-1.67) = 0.9525

Probability decreased

Option D

10)

From z table, P(z<1.88) = 97%

1.88 = (x - 80)/(0.85)

X = 81.598