Question

A government agency reports that 22% of baby boys 6-8 months old in the United States weigh less than 25 pounds. A sample of 147 babies is studied. Use the TI-84 Plus calculator as needed. Round the answer to at least four decimal places. (a) Approximate the probability that more than 40 babies weigh less than 25 pounds. (b) Approximate the probability that 34 or more babies weigh less than 25 pounds. (c) Approximate the probability that the number of babies who weigh less than 25 pounds is between 28 and 38 exclusive. Part 1 of 3 The probability that more than 40 babies weigh less than 25 pounds is .

Answer 1

Answer:

a) 0.0526 = 5.26% probability that more than 40 babies weigh less than 25 pounds.

b) 0.409 = 40.9% probability that 34 or more babies weigh less than 25 pounds.

c) 0.6249 = 62.49% probability that the number of babies who weigh less than 25 pounds is between 28 and 38 exclusive.

Step-by-step explanation:

The binomial approximation to the normal is used to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

A government agency reports that 22% of baby boys 6-8 months old in the United States weigh less than 25 pounds.

This means that $p = 0.22$

A sample of 147 babies is studied.

This means that $n = 147$

Mean and standard deviation:

$\mu = E(X) = np = 147*0.22 = 32.34$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{147*0.22*0.78} = 5.02$

(a) Approximate the probability that more than 40 babies weigh less than 25 pounds.

Using continuity correction, this is P(X > 40 + 0.5) = P(X > 40.5), which is 1 subtracted by the pvalue of Z when X = 40.5. So

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

$Z = \frac{40.5 - 32.34}{5.02}$

$Z = 1.62$

$Z = 1.62$ has a pvalue of 0.9474

1 - 0.9474 = 0.0526

0.0526 = 5.26% probability that more than 40 babies weigh less than 25 pounds.

(b) Approximate the probability that 34 or more babies weigh less than 25 pounds.

Using continuity correction, this is $P(X \geq 34 - 0.5) = P(X \geq 33.5)$, which is 1 subtracted by the pvalue of Z when X = 33.5. So

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

$Z = \frac{33.5 - 32.34}{5.02}$

$Z = 0.23$

$Z = 0.23$ has a pvalue of 0.591

1 - 0.591 = 0.409

0.409 = 40.9% probability that 34 or more babies weigh less than 25 pounds.

(c) Approximate the probability that the number of babies who weigh less than 25 pounds is between 28 and 38 exclusive.

Exclusive means that we dont count 28 and 38, so, using continuity correction, this is $P(28 + 0.5 \leq X \leq 38 - 0.5) = P(28.5 \leq X \leq 37.5)$, which is the pvalue of Z when X = 37.5 subtracted by the pvalue of Z when X = 28.5. So

X = 37.5

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

$Z = \frac{37.5 - 32.34}{5.02}$

$Z = 1.03$

$Z = 1.03$ has a pvalue of 0.8485

X = 28.5

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

$Z = \frac{28.5 - 32.34}{5.02}$

$Z = -0.76$

$Z = -0.76$ has a pvalue of 0.2236

0.8485 - 0.2236 = 0.6249

0.6249 = 62.49% probability that the number of babies who weigh less than 25 pounds is between 28 and 38 exclusive.