Respuesta :
Answer:
[tex] z= \frac{p- \mu_p}{\sigma_p}[/tex]
And the z score for 0.4 is
[tex] z = \frac{0.4-0.4}{\sigma_p} = 0[/tex]
And then the probability desired would be:
[tex] P(p<0.4) = p(z<0) =0.5[/tex]
Step-by-step explanation:
The normal approximation for this case is satisfied since the value for p is near to 0.5 and the sample size is large enough, and we have:
[tex] np = 45*0.4= 18 >10[/tex]
[tex] n(1-p) = 45*0.6= 27 >10[/tex]
For this case we can assume that the population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
Where:
[tex]\mu_{p}= \hat p = 0.4[/tex]
[tex]\sigma_p = \sqrt{\frac{p(1-p)}(n} =\sqrt{\frac{0.4(1-0.4)}(45}= 0.0703[/tex]
And we want to find this probability:
[tex] P(p <0.4)[/tex]
And we can use the z score formula given by:
[tex] z= \frac{p- \mu_p}{\sigma_p}[/tex]
And the z score for 0.4 is
[tex] z = \frac{0.4-0.4}{\sigma_p} = 0[/tex]
And then the probability desired would be:
[tex] P(p<0.4) = p(z<0) =0.5[/tex]
Answer:
The probability that 40% or fewer of the sample are boys P(x ≤ 0.4) = 0.090
Step-by-step explanation:
Normally, the mean number of boys in the population should be given, but in the absence, we take 50% of the population to be boys (since, there are only two possibilities, with close probabilities).
Mean = 0.5
Standard deviation of a sample = √[p(1-p)/n]
p = 0.5
1-p = 0.5
n = sample size = 45
Standard deviation = √[(0.5×0.5)/45] = 0.0745
So, we can now standardize 0.4.
The standardized score is the value minus the mean then divided by the standard deviation.
z = (x - μ)/σ = (0.4 - 0.5)/0.0745 = - 1.34
To determine the probability that 40% or fewer of the sample are boys P(x ≤ 0.4) = P(z ≤ -1.34)
We'll use data from the normal probability table for these probabilities
P(x ≤ 0.4) = P(z ≤ -1.34) = 0.090
