(1/2)+(1/2)<br> lets see who can solve this problem the fastest<br>

The research group asked the following question of individuals who earned in excess of $100,000 per year and those who earned l

Evgen [1.6K]

Answer:

they should

Step-by-step explanation:

5 0

3 years ago

A certain car model has a mean gas mileage of 28 miles per gallon with a standard deviation 5 mpg. Suppose nothing is known abou

professor190 [17]

<span>the probability that the average mileage of the fleet us between 27.2 and 28.8 will be:
27.2<x<28.8
this will be:
P(x<28.8)-P(x<27.2)
To evaluate this we use the z-score formula:
P(x<28.2)=P(z<Z)
Z=(x-</span>μ)/σ
μ-mean
σ-standard deviation
thus
z=(28.2-28)/5=0.04
thus
P(x<28.2)=0.5160

p(x<27.2
z=(27.2-8)/5=-0.16
P(z<-0.16)=0.4364
thus
P(x<28.8)-P(x<27.2)=0.5160-0.4364=0.0796

Answer: 0.0796

3 0

4 years ago

In order to test whether the average waiting time this year differs from last year, a sample of 25 data were collected this year

irina [24]

Answer:

True

Step-by-step explanation:

Confidence interval for banking service in a given confidence level can be calculated as M±ME where

M is the mean banking service in the sample and
ME is the Margin of Error

margin of error (ME) is calculated using the formula

ME= $\frac{t*s}{\sqrt{N} }$ where

t is the corresponding statistic in the given confidence level

s is the standard deviation of the sample(or of the population if it is known)

N is the sample size

t-statistic for 99% confidence level is always bigger than 95% confidence level which makes Margin of Error bigger and thus confidence interval wider.

6 0

4 years ago

canning operation is designed to fill cans with 14.5 oz of ingredients, on average, with a standard deviation of 0.1 oz. Nine ca

Pavlova-9 [17]

Answer:

SE = 0.025

Step-by-step explanation:

We are given;

Sample mean; x¯ = 14.52

Sample standard deviation; s = 0.075

Sample size; n = 9

Now,formula for standard error of sample mean is;