Question

3, The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with a mean of 1263 chips and a standard deviation of 118 chips.

​

​(a) The 27th percentile for the number of chocolate chips in a bag of chocolate chip cookies is_____ chocolate chips.

​(Round to the nearest whole number as​ needed.)

​(b) The number of chocolate chips in a bag that make up the middle 95​% of bags is ___ to ___ chocolate chips.

​(Round to the nearest whole number as needed. Use ascending​ order.)

​(c) The interquartile range of the number of chocolate chips is___.

​(Round to the nearest whole number as​ needed.)

            

4,  Suppose a simple random sample of size n =1000 is obtained from a population whose size is N=1,500,000 and whose population proportion with a specified characteristic is p=0.49. Complete parts​ (a) through​ (c) below.

​(a) Describe the sampling distribution of p^.

A.

Approximately​ normal, mu Subscript ModifyingAbove p with caretμpequals=0.49 and sigma Subscript ModifyingAbove p with caretσpalmost equals≈0.0004

B.

Approximately​ normal, mu Subscript ModifyingAbove p with caretμpequals=0.49 and sigma Subscript ModifyingAbove p with caretσpalmost equals≈0.0158

C.

Approximately​ normal, mu Subscript ModifyingAbove p with caretμpequals=0.49 and sigma Subscript ModifyingAbove p with caretσpalmost equals≈0.0002

​(b) What is the probability of obtaining x=510 or more individuals with the​ characteristic?

​P(x≥510​)equals=nothing ​(Round to four decimal places as​ needed.)

​(c) What is the probability of obtaining x=450 or fewer individuals with the​ characteristic?

​P(x≤450​)=nothing ​(Round to four decimal places as​ needed.)

Answer 1

for normal distribution z score =(X-μ)/σx
here mean=       μ=	1263
std deviation   =σ=	118.000

3)

a)

for 27th percentile critical value of z=		-0.61
therefore corresponding value=mean+z*std deviation=			1191

b)for middle 95% , critical z =1.96

1032 to 1494  chocolate chips.

c)

for  interquartile range  critical z =0.67

1184 to 1342 chocolate chips.

4)

for normal distribution z score =(p̂-p)/σp
here population proportion=     p=	0.4900
sample size       =n=	1000
std error of proportion=σp=√(p*(1-p)/n)=	0.0158

option B is correct

b)

probability =P(X>0.51)=P(Z>(0.51-0.49)/0.016)=P(Z>1.27)=1-P(Z<1.27)=1-0.898=0.1020

c)

probability =P(X<0.45)=(Z<(0.45-0.49)/0.016)=P(Z<-2.5303)=0.0057