Question

In the U.S., 95% of children have received their DTaP vaccine. Calculate the probability that fewer than 700 out of a sample of LaTeX: n = 750n=750 children received their DTaP vaccine.

Answer 1

Answer:

0.0183

Step-by-step explanation:

The percentage of students, who received their DTaP vaccine is 95%.

This implies that:

$p = 0.95$

and

$q = 1 - 0.95 = 0.05$

The standard deviation of the proportion is

$= \sqrt{ \frac{pq}{n} }$

$= \sqrt{ \frac{0.95 \times 0.05}{750} } \\ = 0.00796$

We want to find the probability, that the sample proportion,

$\hat p \leqslant \frac{700}{750} \\ \hat p \leqslant 0.9 \bar3$

$P(\hat p\le0.9\bar3)=P(Z\le \frac{0.9 \bar3 - 0.95 }{0.00796} ) \\ = P(Z\le - 2.09) \\ = 0.0183$

Answer 2

Using the normal approximation to the binomial, it is found that there is a 0.0146 = 1.46% probability that fewer than 700 children received their vaccine.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

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

• It measures how many standard deviations the measure is from the mean.  

• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

• The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with $\mu = np, \sigma = \sqrt{np(1-p)}$.

In this problem:

• 95% of children have received their DTaP vaccine, hence $p = 0.95$

• Sample of 750, hence $n = 750$

For the approximation, the mean and the standard deviation are given by:

$\mu = np = 750(0.95) = 712.5$

$\sigma = \sqrt{np(1 - p)} = \sqrt{750(0.95)(0.05)} = 5.97$

Using continuity correction, the probability that fewer than 700 children received their vaccine is P(X < 700 - 0.5) = P(X < 699.5), which is the p-value of Z when X = 669.5.

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

$Z = \frac{699.5 - 712.5}{5.97}$

$Z = -2.18$

$Z = -2.18$ has a p-value of 0.0146.

0.0146 = 1.46% probability that fewer than 700 children received their vaccine.

A similar problem is given at https://brainly.com/question/14424710