Emetia earns $8.74 per hour plus a gas allowance of $3.50 per day at her job.

An article reports that 1 in 500 people carry the defective gene that causes inherited colon cancer. In a sample of 2000 individ

Tema [17]

Answer:

a) P=0.558

b) P=0.021

Step-by-step explanation:

We can model this random variable as a Poisson distribution with parameter λ=1/500*2000=4.

The approximate distribution of the number who carry this gene in a sample of 2000 individuals is:

$P(x=k)=\frac{\lambda^ke^{-\lambda}}{k!} =\frac{4^ke^{-4}}{k!}$

a) We can calculate that the approximate probability that between 4 and 9 (inclusive) as:

$P(4\leq x\leq 9)=\sum_{k=4}^9P(k)\\\\\\ P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\P(9)=4^{9} \cdot e^{-4}/9!=262144*0.0183/362880=0.013\\\\\\$

$P(4\leq x\leq 9)=\sum_{k=4}^9P(k)=0.195+0.156+0.104+0.060+0.030+0.013=0.558$

b) The approximate probability that at least 9 carry the gene is:

$P(x\geq9)=1-P(x\leq 8)\\\\\\$

$P(0)=4^{0} \cdot e^{-4}/0!=1*0.0183/1=0.018\\\\P(1)=4^{1} \cdot e^{-4}/1!=4*0.0183/1=0.073\\\\P(2)=4^{2} \cdot e^{-4}/2!=16*0.0183/2=0.147\\\\P(3)=4^{3} \cdot e^{-4}/3!=64*0.0183/6=0.195\\\\P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\$

$P(x\geq9)=1-P(x\leq 8)\\\\P(x\geq9)=1-(0.018+0.073+0.147+0.195+0.195+0.156+0.104+0.060+0.030)\\\\P(x\geq9)=1-0.979=0.021$

8 0

2 years ago

1. Which of the following is not an equation?

OlgaM077 [116]

Answer:

2. is not an equation

Step-by-step explanation:

An equation always has an equal sign. An expression doesn't

5 0

2 years ago

Part A- solve for x<br> Part B- solve for y<br> Part C- find the perimeter<br> (photo included)

lidiya [134]

Part A: x=3
Part B: y=3

8 0

3 years ago

Hello hope everyone's Day is good if so tell me why or why not

Aliun [14]

Answer:

Well....

A Good Day, Cause Both mY parents are On Duty and Im free, Alone at home!!

3 0

3 years ago

Read 2 more answers

The average daily volume of a computer stock in 2011 was million shares, according to a reliable source. A stock analyst believ

igomit [66]

Complete question :

The average daily volume of a computer stock in 2011 was p = 35.1 million shares, according to a reliable source. A stock analyst believes that the stock volume in 2014 is different from the 2011 level. Based on a random sample of 40 trading days in 2014, he finds the sample mean to be 30.9 million shares, with a standard deviation of s = 11.8 million shares. Test the hypotheses by constructing a 95% confidence interval. Complete parts (a) through (c) below. State the hypotheses for the test. Construct a 95% confidence interval about the sample mean of stocks traded in 2014.

Answer:

H0 : μ = 35.1 ;

H1 : μ < 35.1 ;

(26.488 ; 35.312)

Step-by-step explanation:

The hypothesis :

H0 : μ = 35.1

H1 : μ < 35.1

The confidence interval :

Xbar ± Margin of error

Xbar = 30.9

Margin of Error = Zcritical * s/sqrt(n)

Zcritical at 95% = 1.96

Margin of Error = 1.96 * (11.8/sqrt(40))

Margin of Error = 4.412

Lower boundary :

30.9 - 4.412 = 26.488

Upper boundary :

30.9 + 4.412 = 35.312

Confidence interval = (26.488 ; 35.312)

Since the population mean value exists within the interval, the we fail to reject the Null.

5 0

2 years ago