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.

Answer 1

1)

n=

500

p=

0.0500

here mean of distribution=μ=np=		25.00
and standard deviation σ=sqrt(np(1-p))=		4.87
for normal distribution z score =(X-μ)/σx

therefore from normal approximation of binomial distribution and continuity correction:

probability of winning 30 times :

probability =P(29.5<X<30.5)=P((29.5-25)/4.873)<Z<(30.5-25)/4.873)=P(0.92<Z<1.13)=0.8708-0.8212=0.0496

2)

point estimate 'x̄=	7.500
sample size    n=	1500.00
std deviation σ=	0.800
std error ='σx=σ/√n=	0.0207

for 95 % CI value of z=		1.960
margin of error E=z*std error =		0.040
lower bound=sample mean-E=		7.460
Upper bound=sample mean+E=		7.540

from above 95% confidence interval for population mean =(7.46,7.54)