Assume that the heights of women are normally distributed with a mean of 63.6 in

Question

Assume that the heights of women are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. The U.S. Army requires that the heights of women be between 58 and 80 inches. If 200 women want to enlist in the U.S. Army, how many would you expect to meet the height requirements? Find the equation of the line of best fit from the data in the table. Round the slope and the y-intercept to the the indicated probability. If P(A) = 0.2, P(A or B) = 0.5, and P(A and B) = 0.3, find P(B). the indicated expected frequency. A researcher wants to determine whether the number of minutes adults spend online per day is related to gender. A random sample of 315 adults was selected and the results are shown below. Find the expected frequency for the cell E_2,2 to determine if there is enough evidence to conclude that the number of minutes spent online per day is related to gender. Round to the nearest tenth if necessary. Marbles are drawn without replacement from a box with 3 white, 2 green, 2 red, and 1 blue marble. Find the prob Both marbles are white. 3/8 3/32 3/28 9/56

Explanation / Answer

22.

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    58      
x2 = upper bound =    80      
u = mean =    63.6      
          
s = standard deviation =    2.5      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -2.24      
z2 = upper z score = (x2 - u) / s =    6.56      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.012545461      
P(z < z2) =    1      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.987454539      

Hence, around 0.98745*200 = 197.49 will meet the requirements. [ANSWER: 197.49]

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