Question

Question 1- Suppose that one of every ten cereal boxes has a prize, Out of a shipment of n = 400 cereal boxes, find the probability that there are somewhere between 40 and 46 boxes with prizes. (Hint:use the dishonest coin principle with p = 0.1 to solve for the mean and standard deviation. Remember the probability is a decimal number, round the answer to the nearest hundredth.)

Question 2- A normal distribution has a mean of 30 cm and 47.5% of the data is between 24 and 30 cm. What is the standard deviation. (Put in the number only.)

Answer 1

In Q1. CLT is used which is mentioned in detail in second question.

Dishonest-coin principle:

Let X denote the number of heads in n tosses of a coin. Suppose n ≥ 30 ( n is large). Let p denote the probability of heads on each toss. Then X has an approximately normal distribution with mean µ = n p and variance σ² = n p (1 − p).

CODES:

n=400
p=0.1
pbinom(46,n,p)-pbinom(40,n,p)

ex=n*p
ex
varx=n*p*(1-p)
varx
a=(46-ex)/sqrt(varx)
a
b=(40-ex)/sqrt(varx)
b
pnorm(a)-pnorm(b)

#Codes alongwith outputs:
#Q1.
> n=400
> p=0.1
> pbinom(46,n,p)-pbinom(40,n,p)
[1] 0.3180649
>
> ex=n*p
> ex
[1] 40
> varx=n*p*(1-p)
> varx
[1] 36
> a=(46-ex)/sqrt(varx)
> a
[1] 1
> b=(40-ex)/sqrt(varx)
> b
[1] 0
> pnorm(a)-pnorm(b)
[1] 0.3413447

#Q2.
> pnorm(0)
[1] 0.5
> pnorm(0)-47.5/100
[1] 0.025
> qnorm(0.025)
[1] -1.959964
> -6/qnorm(0.025)
[1] 3.061281
> pnorm(30,30,3.061281)-pnorm(24,30,3.061281)
[1] 0.475