Question

Estimate the probability of getting exactly 43 boys in 90 births. estimate the indicated probability by using the normal distribution as an approximation to the binomial distribution. round to four decimal places.

Answer 1

The probability of getting exactly 43 boys is given by:
P(X = 43) = P(X ≤ 43) - P(X ≤ 42)

First, let's find P(X ≤ 43). We know n = 90 and the probability of getting a boy is p = 0.50
Calculate the mean: μ = n · p = 90 · 0.5 = 45
Calculate the standard deviation: σ = √[n · p · (1 - p)] = √(90 · 0.5 · 0.5) = 4.743
Calcuate the z-score: z = (X - μ) / σ = (43 - 45) / 4.743 = -0.42
Now, look at a z-score table in order to find to what probability this value corresponds: P(z ≤ -0.42) = 0.3372 = P(X ≤ 43)

Now, repeat the calculations for X = 42: 
μ and  σ stay the same,
z = (X - μ) / σ = (42 - 45) / 4.743 = -0.63
P(z ≤ -0.63) = 0.2643 = P(X ≤ 42)

Now, find the difference:
P(X = 43) = P(X ≤ 43) - P(X ≤ 42) = 0.3372 - 0.2643 = 0.0729

You have around 7.3% of probability to get exactly 43 boys in 90 births.

Answer 2

Answer:

There is a 7.67% chance of getting exactly 43 boys in 90 births.

Step-by-step explanation:

Hint- We have to calculate mean(μ) and standard deviation(σ) from binomial distribution and then calculate the z score for normal distribution. After then we can get the probability.

For Binomial distribution,

n = sample size = 90

p = probability of event of getting a boy = 0.5 (as we have only two choices, boy and girl)

Putting these in the formulae,

$\mu=n\cdot p=90(0.5) = 45$

$\sigma=\sqrt{n\cdot p\cdot (1-p)}=\sqrt{90\times 0.5 \times 0.5} =4.7434$

Now we have to convert the z-values using the continuity correction as the binomial is a discrete distribution. Because as in the question the probability of getting exactly 43 boys in 90 births is asked. Number of boys can not be in fraction, so it is discrete.

In normal distribution,

$Z=\dfrac{X-\mu}{\sigma}$

So,

$P(X = 43) = P[(\dfrac{42.5 - 45}{4.7434}) < z <(\dfrac{43.5 - 45}{4.7434})]$

$= P[-0.527 < z < -0.316]$

Using the standard normal table,

$P[-0.527 < z < -0.316]=0.3782-0.3015=0.0767=7.67\%$