The height (in inches) of men at uh is assumed to have a normal distribution with a standard deviation of 3.6 inches. the height (in inches) of women at uh is also assumed to have a normal distribution with a standard deviation of 2.9 inches. a random sample of 49 men and 38 women yielded respective means of 68.3 inches and 64.6 inches. find the 90% confidence interval for the difference in the heights of men at uh and women at uh

The required 90% confidence interval would be (2.55,4.85).

Step-by-step explanation:

Since we have given that

Standard deviation of men = 3.6 inches

Standard deviation of women = 2.9 inches

Average of men = 68.3 inches

Average of women = 64.6 inches

Number of men = 49

Number of women = 38

α = 0.10

So, the confidence interval would be

Hence, the required 90% confidence interval would be (2.55,4.85).

4√5 + 4√10

Step-by-step explanation:

The perimeter is the sum of all the sides.

Using the coordinates given to us, we can find the sides of the quadrilateral.

d(ab) = √[(x2 - x1)² + (y2 - y1)²]

Given that, we apply that to each and every of the side, and thus we have

d(le) = √[(3 --3)² + (3 - 1)²]

d(le) = √[(6²) + (2)²]

d(le) = √(36 + 4)

d(le) = √40

d(ea) = √[(5 - 3)² + (7 - 3)²]

d(ea) = √[(2)² + (4²)]

d(ea) = √(4 + 16)

d(ea) = √20

d(ap) = √[(-1 - 5)² + (5 - 7)²]

d(ap) = √[-6)² + (-2)²]

d(ap) = √(36 + 4)

d(ap) = √40

d(pl) = √[(-1 --3)² + (5 - 1)²]

d(pl) = √[(-2)² + (4)²]

d(pl) = √(4 + 16)

d(pl) = √20

Perimeter of the quadrilateral is then

d(le) + d(ea) + d(ap) + d(pl)

√40 + √20 + √40 + √20

4√5 + 4√10