Question

1. Use the​ 68-95-99.7 Rule to approximate the probability rather than using technology to find the values more precisely. The daily closing price of a stock​ (in $) is well modeled by a Normal model with mean ​$158.14 and standard deviation ​$4.64. According to this​ model, what cutoff​ value(s) of price would separate the following​ percentage? ​

a) highest 0.15​%

​b) highest ​50%

​c) middle 95​%

​d) highest 2.5​%

2. In the last quarter of​ 2007, a group of 64 mutual funds had a mean return of 4.6​% with a standard deviation of 2.5​%. Consider the Normal model ​N(0.046​, 0.025​) for the returns of these mutual funds.

​a) What value represents the 40th percentile of these​ returns?

​b) What value represents the 99th​ percentile? ​

c)​ What's the​ IQR, or interquartile​ range, of the quarterly returns for this group of​ funds?

Answer 1

a) highest 0.15% will fall at 3 standard deviaiton above mean: corresponding value =158.14+3*4.64 =172.06

b) highest 505 will be at mean, corresponding value =158.14

c) middle 95% are 2 standard deviation from mean : corresponding values =158.14 -/+ 2*4.64 =148.86 to 167.42

d) highest 2.5​% will fall at 2 standard deviation above mean: corresponding value =158.14+2*4.64 =167.42

2)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	0.046
std deviation   =σ=	0.025

a)

for 40th percentile critical value of z=		-0.25
therefore corresponding value=mean+z*std deviation=			0.0398 or 3.98%

b)

for 99th percentile critical value of z=		2.33
therefore corresponding value=mean+z*std deviation=			0.1043 or 10.43%

c)

for 75th percentile critical value of z=		0.67
therefore Q1 =mean+z*std deviation=			0.0628

for 25th percentile critical value of z=		-0.67
therefore Q2=mean+z*std deviation=			0.0293

​ IQR, or interquartile​ range =Q3-Q1 =0.0628-0.0293 =0.0335 or 3.35%