Question

The amount of coffee that people drink per day is normally distributed with a mean of 17 ounces and a standard deviation of 4 ounces. 15 randomly selected people are surveyed. Round all answers to 4 decimal places where possible.
a) What is the distribution of XX? XX ~ N(,)
b) What is the distribution of ¯xx¯? ¯xx¯ ~ N(,)
c) What is the probability that one randomly selected person drinks between 15.5 and 18 ounces of coffee per day?
d) For the 15 people, find the probability that the average coffee consumption is between 15.5 and 18 ounces of coffee per day.
e) For part d), is the assumption that the distribution is normal necessary? YesNo
f) Find the IQR for the average of 15 coffee drinkers.
Q1 = ounces
Q3 = ounces
IQR: ounces

Answer 1

Answer:

Step-by-step explanation:

(a)

The distribution of X is Normal Distribution with mean $= \mu =17$ and Variance $= \sigma^{2} = 16 \ i.e., X \sim N (17, 16),$

(b)

The distribution of $\bar{x}$ is Normal Distribution with mean $= \mu =17$ and Variance = $\sigma^{2}/n = 16/15= 1.0667$.i.e., $\bar{x}\sim N(17,1.0667)$

c)

To find P(15.5 < X < 18):

Case 1: For X from 15.5 to mid value:

Z = (15.5 - 17)/4 = - 0.375

Table of Area Under Standard Normal Curve gives area = 0.1480

Case 2: For X from mid value to 18:

Z = (18 - 17)/4 = 0.25

Table of Area Under Standard Normal Curve gives area = 0.0987

So,

P(15.5 < X< 18) = 0.1480 +0.0987 = 0.2467

So,

Answer is:

0.2467

(d)

$SE = \sigma/\sqrt{n}\\\\= 4/\sqrt{15}$

= 1.0328

To find $P(15.5 < \bar{x}< 18):$

Case 1: For $\bar{x}$ from 15.5 to mid value:

Z = (15.5 - 17)/1.0328 = - 1.4524

Table of Area Under Standard Normal Curve gives area = 0.4265

Case 2: For X from mid value to 18:

Z = (18 - 17)/1.0328 = 0.9682

Table of Area Under Standard Normal Curve gives area = 0.3340

So,

$P(15.5 < \bar{x}< 18) = 0.4265 + 0.3340 = 0.7605$

So,

Answer is:

0.7605

(e)

Correct option:

No

because Population SD is provided.

(f)

(i)

Q1 is given by:

$- 0.6745 = (\bar{x} - 17)/1.0328$

So,

X = 17 - (0.6745 * 1.0328) = 17 - 0.6966 = 16.3034

So,

Q1 = 16.3034

(ii)

Q3 is given by:

$0.6745 = (\bar{x} - 17)/1.0328$

So,

X = 17 + (0.6745 * 1.0328) = 17 + 0.6966 = 17.6966

So,

Q3= 17.6966

(iii)

IQR = Q3 - Q1 = 17.6966 - 16.3034 = 1.3932

So

Answers are:

Q1 = 16.3034 ounces

Q3 = 17.6966 Ounces

IQR = 1.3932 Ounces