1.Assume that men’s weights are normally distributed with a mean given by  = 172lb and a standard deviation given by  =29lb. Using the Central Limit Theorem to solve the following exercises(1)If 36 men are randomly selected, find the probability that they have a mean weight greater than 160lb.(2)If 81 men randomly selected, find the probability that they have a mean weight between 170lb and 175lb.

1) 0.99348

2) 0.55668

Step-by-step explanation:

Assume that men’s weights are normally distributed with a mean given by  = 172lb and a standard deviation given by  =29lb. Using the Central Limit Theorem to solve the following exercises

When given a random number of samples, we use the z score formula:

z-score is z = (x-μ)/σ/√n where

x is the raw score

μ is the population mean

σ is the population standard deviation.

(1) If 36 men are randomly selected, find the probability that they have a mean weight greater than 160lb.

For x > 160 lb

z = 160 - 172/29/√36

z = 160 - 172/29/6

z = -2.48276

Probability value from Z-Table:

P(x<160) = 0.0065185

P(x>160) = 1 - P(x<160) = 0.99348

(2) If 81 men randomly selected, find the probability that they have a mean weight between 170lb and 175lb.

For x = 170 lb

z = 170 - 172/29/√81

z = 170 - 172/29/9

z = -0.62069

Probability value from Z-Table:

P(x = 170) = 0.2674

For x = 175 lb

z = 175 - 172/29/√36

z = 175- 172/29/6

z = 0.93103

Probability value from Z-Table:

P(x = 175) = 0.82408

The probability that they have a mean weight between 170lb and 175lb is calculated as:

P(x = 175) - P(x = 170)

0.82408 - 0.2674

= 0.55668

d) 30 units^2

area of rectangle = l × b

we know,

length = 10 unit

breadth = 3 unit

applying the formula,

= l × b

= 10 × 3

= 30 unit^2

happy to

pls mark as brainliest

n=9

step-by-step explanation:

7 times a number = 7n

decreased by 13 = 7n-13

same as is =

5 times the number = 5n

increased by 5 = 5n+5

7n-13 = 5n+5

2n=18

n=9