Question

The probability of winning on a lot machine is 5%. If a person plays the machine 500 times, find the probability of winning 30 times. Use the normal approximation to the binomial distribution.

A travel survey of 1500 Americans reported an average of 7.5 nights stayed when they went on vacation. Find a point estimate of the population mean. If we can assume the population standard deviation is 0.8 night, find the 95% confidence interval for the true mean.

SHOW CLEAR AND EASY WORK TO FOLLOW PLEASE

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 7.5

Population standard deviation =    = 0.8

Sample size = n = 1500

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_0.025 = 1.96

Margin of error = E = Z/2 * ( /n)

= 1.96 * ( 0.8 /  1500 )

= 0.04

At 95% confidence interval estimate of the population mean is,

- E < < + E

7.5 - 0.04 <   < 7.5 + 0.04

( 7.46 <   < 7.54 )