Question

Suppose babies born in a large hospital have a mean weight of 3181 grams, and a standard deviation of 526 grams. If 97 babies are sampled at random from the hospital, what is the probability that the mean weight of the sample babies would differ from the population mean by greater than 53 grams? Round your answer to four decimal places.

Answer 1

Answer:

0.3222 = 32.22% probability that the mean weight of the sample babies would differ from the population mean by greater than 53 grams

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 3181, \sigma = 526, n = 97, s = \frac{526}{\sqrt{97}} = 53.41$

Probability that the mean weight of the sample babies would differ from the population mean by greater than 53 grams?

Two cases, the mean being lesser than 3181 - 53 = 3128 grams or it being greater than 3181 + 53 = 3234.

Lesser than 3128

pvalue of Z when X = 3128.

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

By the Central Limit Theorem

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

$Z = \frac{3128 - 3181}{53.41}$

$Z = -0.99$

$Z = -0.99$ has a pvalue of 0.1611

Greater than 3234

1 subtracted by the pvalue of Z when X = 3234

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

$Z = \frac{3234 - 3181}{53.41}$

$Z = 0.99$

$Z = 0.99$ has a pvalue of 0.8389

1 - 0.8389 = 0.1611

2*0.1611 = 0.3222

0.3222 = 32.22% probability that the mean weight of the sample babies would differ from the population mean by greater than 53 grams

Answer 2

Answer:

$P(\bar X <3181-53=3128) \cup P(\bar x>3181+53=3234)$

$P(\bar X < 3128)$

And we can use the z score given by:

$z = \frac{\bar X- \mu }{\frac{\sigma}{\sqrt{n}}}$

The z score for 3128 is:

$z= \frac{3128-3181}{\frac{526}{\sqrt{97}}}= -0.992$

And we can use the normal standard table or excel and we got:

$P(Z<-0.992)= 0.1606$

For the other probability we have:

$P(\bar X < 3234)$

And we can use the z score given by:

$z = \frac{\bar X- \mu }{\frac{\sigma}{\sqrt{n}}}$

The z score for 3234 is:

$z= \frac{3234-3181}{\frac{526}{\sqrt{97}}}= 0.992$

And we can use the complement rule, the normal standard table or excel and we got:

$P(Z>0.992)=1-P(Z<0.992)=1- 0.8394=0.1606$

And the final probability would be 0.1606+0.1606= 0.3212

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variable of interest and we know that:

$E(X)= 3181 , Sd(X) = 526$

We select a random sample of n=97 nails. That represent the sample size.  

And we want to calculate the probability that the mean weight of the sample babies would differ from the population mean by greater than 53 grams

$P(\bar X <3181-53=3128) \cup P(\bar x>3181+53=3234)$

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

We can find the individual probabilities and we got:

$P(\bar X < 3128)$

And we can use the z score given by:

$z = \frac{\bar X- \mu }{\frac{\sigma}{\sqrt{n}}}$

The z score for 3128 is:

$z= \frac{3128-3181}{\frac{526}{\sqrt{97}}}= -0.992$

And we can use the normal standard table or excel and we got:

$P(Z<-0.992)= 0.1606$

For the other probability we have:

$P(\bar X < 3234)$

And we can use the z score given by:

$z = \frac{\bar X- \mu }{\frac{\sigma}{\sqrt{n}}}$

The z score for 3234 is:

$z= \frac{3234-3181}{\frac{526}{\sqrt{97}}}= 0.992$

And we can use the complement rule, the normal standard table or excel and we got:

$P(Z>0.992)=1-P(Z<0.992)=1- 0.8394=0.1606$

And the final probability would be 0.1606+0.1606= 0.3212