Question

1) A population of values has a distribution with μ=6μ=6 and σ=23.1σ=23.1. You intend to draw a random sample of size n=90n=90.

According to the Central Limit Theorem:

(a) What is the mean of the distribution of sample means?
μ¯x=μx¯=  

(b) What is the standard deviation of the distribution of sample means?
(Report answer accurate to 2 decimal places.)
σ¯x=σx¯=  

(c) In a random sample of n=90, what is the probability that its sample mean is more than 4.2? Round to three decimal places.



(d) In a random sample of n=90, what is the probability that its sample mean is less than 9.9? Give your answer to three decimal places.

2) In a survey, 47 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $67 and standard deviation of $9.

Calculate, state, and interpret a 95% confidence interval to estimate the mean amount of money parents spend on their child's birthday gift. Round to the nearest 100th where necessary.

Answer 1

1)Solution:

Given that,

μ= 6,σ = 23.1,n= 90

By using the definition of central limit theorem,

For a random sample of size n >=30 , sample mean is equal to the population mean

i.e. μx̄ = μ and

sample standard deviation can be calculated as

σx̄=σ/(√n)

Therefore,for a random sample of size n=90 sample mean is equal to population mean

a) μx̄ = μ= 6

The mean of the sampling distribution is μx̄ = 6

And standard deviation is

b) σx̄=σ/(√n)

=23.1 /(√90)

= 2.4349

= 2.43 .... upto 2 decimal place.

The standard deviation of the distribution of sample mean is σx̄= 2.43

c)P(x̄>4.2)=1- P(x̄<=4.2)

= 1-p{[(x̄- μx̄ )/σx̄]<=[(4.2 - 6)/2.43]}

=1-P(z<= -0.74)

=1- 0.2296

= 0.7704

P(x̄>4.2)=0.770... upto 3 decimal place.

d)P(x̄<9.9)= p{[(x̄- μx̄ )/σx̄]<[(9.9 - 6)/2.43]}

=P(z< 1.60)

=0.9452

P(x̄<9.9)=0.945 ... upto 3 decimal place.

2)Solution:

Given that, n=47, x̄= $67

σ= $9

(1–α)%=95%

α=0.05

α/2 =0.025

Zα/2 =1.96 ....from standard normal table.

Margin of error=E=Zα/2 ×(σ/✓n)

=1.96 ×(9/✓47)

=1.96 × 1.3128

=2.5731

Margin of error=E=2.5731

95% confidence interval to estimate the mean amount of money parents spend on their child's birthday gift is given as,

x̄± Margin of error

(67 - 2.5731,67 + 2.5731)

(64.4269,69.5731)

We are 95% confident that the mean amount of money parents spend on their child's birthday gift is lies between confidence interval ($64.4269,$69.5731)