Question

A fair coin is tossed 10 times and the number of heads is counted. Complete parts​ (a) through​ (d).

a. Use the binomial distribution to find the probability of getting 5 heads.

​(Round to four decimal places as​ needed.)

b. Use the binomial distribution to find the probability of getting at least 5 heads.

​(Round to four decimal places as​ needed.)

c. Use the binomial distribution to find the probability of getting 5 to 7 heads.

​(Round to four decimal places as​ needed.)

d. Use the normal distribution to approximate the probability of getting at least 5 heads. (2 points)

​(Round to four decimal places as​ needed.)

Answer 1

Solution:

n = 10

Let X be the number of heads is counted.

p be the probability of success of heads

p = 0.5

X follows Binomial(n = 10 , p = 0.5)

Using binomial probability formula ,

P(X = x) = (_n C _x) * p^x * (1 - p)^{n - x} ; x = 0 ,1 , 2 , ....., n

a)

P[getting 5 heads]

= P[X = 5]

=  (₁₀ C ₅) * 0.5⁵ * (1 - 0.5)^{10 - 5}

= 0.2461

P[getting 5 heads] = 0.2461

b)

P[getting at least 5 heads]

= P[X $\geq$ 5]

= 1 - { P[X < 5] }

= 1 - { P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) }

= 1 - {  (₁₀ C ₀) * 0.5⁰ * (1 - 0.5)^{10 - 0} +  (₁₀ C ₁) * 0.5¹ * (1 - 0.5)^{10 - 1} +  (₁₀ C ₂) * 0.5² * (1 - 0.5)^{10 - 2} +  (₁₀ C ₃) * 0.5³ * (1 - 0.5)^{10 - 3} +  (₁₀ C ₄) * 0.5⁴ * (1 - 0.5)^{10 - 4} }

= 1 - { 0.0009765625+0.009765625+0.0439453125+0.1171875+0.205078125 }

= 1 - 0.376953125

= 0.6230

P[getting at least 5 heads] =  0.6230

c)

P[getting 5 to 7 heads]

= P(X = 5) + P(X = 6) + P(X = 7)

=  (₁₀ C ₄) * 0.5⁴ * (1 - 0.5)^{10 - 4} +   (₁₀ C ₆) * 0.5⁶ * (1 - 0.5)^{10 - 6} +   (₁₀ C ₇) * 0.5⁷ * (1 - 0.5)^{10 - 7} ​​​​​​​+

= 0.24609375 + 0.205078125 + 0.1171875

= 0.5684

P[getting 5 to 7 heads] =  0.5684

d)

According to normal approximation binomial,

X $\rightarrow$ Normal

Mean = $\mu$ = n*p = 10 * 0.5 = 5

Standard deviation = $\sigma$ =$\sqrt{}$n*p*(1-p) = $\sqrt{}$ [10*0.5*0.5]  = 1.58113883008

Now ,

P[getting at least 5 heads.]

= P[X $\geq$ 5]

Using continuity correction ,

= P[X > 4.5]

= P[(X - $url=https://latex.codecogs.com/gif.latex?%5Cmu&t=1596163916&ymreqid=f4764063-0d40-e2b0-1cde-f00053012f00&sig=PrkMvSl1tOa6cBMo7G19mg--~D$ )/$url=https://latex.codecogs.com/gif.latex?%5Csigma&t=1596163916&ymreqid=f4764063-0d40-e2b0-1cde-f00053012f00&sig=UUwRWH102YhAW8Ogu4PJHg--~D$ >  (4.5 - $url=https://latex.codecogs.com/gif.latex?%5Cmu&t=1596163916&ymreqid=f4764063-0d40-e2b0-1cde-f00053012f00&sig=PrkMvSl1tOa6cBMo7G19mg--~D$ )/$url=https://latex.codecogs.com/gif.latex?%5Csigma&t=1596163916&ymreqid=f4764063-0d40-e2b0-1cde-f00053012f00&sig=UUwRWH102YhAW8Ogu4PJHg--~D$]

= P[Z > (4.5 - 5)/1.58113883008]

= P[Z > -0.316]

= 1 - P[Z < -0.316]

= 1 - 0.3760 ( use z table)

= 0.6240

P[getting at least 5 heads] =  0.6240