Find the necessary confidence interval for the binomial proportion p. (round your answers to three decimal places.) a 90% confidence interval for p, based on a random sample of n = 400 observations from a binomial population with x = 358 successes.

[0.875;0.925]

Step-by-step explanation:

Hello!

You have a random sample of n= 400 from a binomial population with x= 358 success.

Your variable is distributed X~Bi(n;ρ)

Since the sample is large enough you can apply the Central Limit Teorem and approximate the distribution of the sample proportion to normal

^ρ≈N(ρ;(ρ(1-ρ))/n)

And the standarization is

Z= ^ρ-ρ  ≈N(0;1)

√(ρ(1-ρ)/n)

The formula to estimate the population proportion with a Confidence Interval is

[^ρ ± *√(^ρ(1-^ρ)/n)]

The sample proportion is calculated with the following formula:

^ρ= x/n = 358/400 = 0.895 ≅ 0.90

And the Z-value is ≅ 1.65

[0.90 ± 1.65 * √((0.90*0.10)/400)]

[0.875;0.925]

I hope you have a SUPER day!

.632

step-by-step explanation:

6 * 1/10 = 3/5

3*0.01 = 0.03

3/5 + 0.03 = .63

2*0.002 = .002

.63 + 0.002 = .632

sas (side angle side)

aa(angle angle)

sss(side side side)

sas you have to find two sides with an angle in between

aa you find two angles because it will always add up to 180

sss find all sides

step-by-step explanation: