Suppose the heights of 18-year-old men are approximately normally distributed, with mean 70 inches and standard deviation 6 inches.
(a) What is the probability that an 18-year-old man selected at random is between 69 and 71 inches tall? (Round your answer to four decimal places.)
(b) If a random sample of twenty 18-year-old men is selected, what is the probability that the mean height x is between 69 and 71 inches? (Round your answer to four decimal places.)
(c) Compare your answers to parts (a) and (b). Is the probability in part (b) much higher? Why would you expect this?
-The probability in part (b) is much lower because the standard deviation is smaller for the x distribution.
-The probability in part (b) is much higher because the mean is larger for the x distribution.
-The probability in part (b) is much higher because the standard deviation is larger for the x distribution.
-The probability in part (b) is much higher because the mean is smaller for the x distribution. The probability in part (b) is much higher because the standard deviation is smaller for the x distribution.
(A) mean 70 inches and standard deviation 6 inches.
Using normalcdf(lower bound, upper bound, mean, standard deviation)
setting lower bound= 69, upper bound= 71, mean= 70, standard deviation = 6
this implies
=normalcdf(69,71,70,6)
= 0.1324
(B) sample standard deviation = (standard deviation)/(sqrt{n})
= 6/sqrt{20}
mean 70 inches and standard deviation 6/sqrt(20) inches.
Using normalcdf(lower bound, upper bound, mean, standard deviation)
setting lower bound= 69, upper bound= 71, mean= 70, standard deviation = 6/sqrt{20}
this implies
=normalcdf(69,71,70,6/sqrt{20})
= 0.5439
(C) it is clear from the above calculations that the probability in part b is much higher than the probability in part a because the standard deviation is smaller
so, option D is correct
The probability in part (b) is much higher because the standard deviation is smaller for the x distribution.
Get Answers For Free
Most questions answered within 1 hours.