Adult entrance fees to amusement parks in the united states are normally distributed with a population standard deviation of 2.5 dollars and an unknown population mean. a random sample of 22 entrance fees at different amusement parks is taken and results in a sample mean of 61 dollars. find the margin of error for a 99% confidence interval for the population mean.

 1.509

Step-by-step explanation:

The formula of Margin of Error for (n<30):-

Given : Sample size : n= 22

Level of confidence = 0.99

Significance level :

Using the t-distribution table ,

Critical value :

Standard deviation:

Then, we have

Hence, the margin of error for a 99% confidence interval for the population mean =1.509

62.5 paisa

step-by-step explanation:

4000*70 = 280000

280000-(2000*90) = 100000

4000-2400 = 1600

100000/1600 = 62.5

( give brainliest)

25 questions , because they start with 100.  

350-100=250 .  

. 250/10=25

step-by-step explanation:

first take off 100, then what x 10 = 250? that is your answer. (25)