A large population of variable x is characterized by its known mean value of 6.1 units and a standard deviation of 1.0 units and a normal distribution. Determine the range of values containing 70% of the population of x

-6.134 to +6.134

Step-by-step explanation:

given that a large population of variable x is characterized by its known mean value of 6.1 units and a standard deviation of 1.0 units and a normal distribution

X is Normal with mean  =6.1 and std dev = 1 unit

We are to determine the range of values containing 70% of the population of x

We know that normal distribution curve is bell shaped symmetrical about the mean.

So to find 70% range we can use 35% on either side of the mean

Using std normal distribution table the value of z for which probability from 0 to z is 0.35 is 1.034

Hence corresponding x value is

i.e. 70% values lie between

-6.134 to +6.134

5

step-by-step explanation:

answer:   28.56 feet (approx)

step-by-step explanation:

since, triangle abc has measure of angle c equal to 55 degrees, measure of angle abc equal to 90 degrees, and length of bc equal to 20 feet.

we have to find out the measurement of segment ab.

therefore, tan 55° =

but bc = 20 feet.

⇒ tan 55 ° =  

⇒ ab = 20 × tan 55 °

⇒ ab = 20 × 1.42814800674

⇒ ab = 28.5629601348   ≈ 28.56 feet