Heights of adult women are normally distributed with a mean of 64 inches with a standard deviation of 2.5 inches. Between what two heights would approximately 95% of the data be contained? OA. 58.00 and 68.00 inches OB. 61.50 and 66.50 inches O C. 58.50 and 71.50 inches O D. 59.00 and 69.00 inches

Previous Question Researchers measured the data speeds for a particular smartphone carier at 50 airports. The highest speed measured was T76 Mbps. The complete list of 50 data speeds has a mean of x 17.47 Mbps and a standard deviation of s 32.99 Mbps. a. What is the dilference between camer's highest data speed and the mean of all 50 data speeds? b. How many standard deviations is that (the difference found in part (a)? c Convert the camer's highest data speed to az score. d. If we consider data speeds that convert to z scores between-2 and 2 to be neither significantly low nor significantly high, is the camier's highest data speed significant? a. The difference is Mbps (Type an integer or a decimal Do not round) b. The dilference is standard deviations (Round to two decimal places as needed.) c The z score is z (Round to two decimal places as needed.) d. The camer's highest data speed is

Assume that males have a mean hieght of 69.1 in. with a standard deviation of 2.9 in., and females have a mean height of 63.7 in. with a standard deviation of 2.7 in. Based on Z scores, would a 5'4" male or a 4'10" female be taller compared to the rest of their gender?

The data set of size 40 has mean 15 and standard deviation 2. Approximately how many observations lie between 13 and 17?