The narrator of Fever 1793 helps readers imagine Philadelphia summers by describing

According to the National Vital Statistics, full-term babies' birth weights are Normally distributed with a mean of 7.5 pounds a

Sav [38]

Answer:

68.26% probability that a randomly selected full-term pregnancy baby's birth weight is between 6.4 and 8.6 pounds

Step-by-step explanation:

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.

In this problem, we have that:

$\mu = 7.5, \sigma = 1.1$

What is the probability that a randomly selected full-term pregnancy baby's birth weight is between 6.4 and 8.6 pounds

This is the pvalue of Z when X = 8.6 subtracted by the pvalue of Z when X = 6.4. So

X = 8.6

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

$Z = \frac{8.6 - 7.5}{1.1}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

X = 6.4

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

$Z = \frac{6.4 - 7.5}{1.1}$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

68.26% probability that a randomly selected full-term pregnancy baby's birth weight is between 6.4 and 8.6 pounds

6 0

3 years ago

Suppose that a recent article stated that the mean time spent in jail by a first-time convicted burglar is 2.5 years. A study wa

4vir4ik [10]

Using the z-distribution, it is found that since the test statistic is greater than the critical value, it can be concluded that the mean length of jail time has increased.

At the null hypothesis, it is tested if the mean length of jail time is still of 2.5 years, that is:

$H_0: \mu = 2.5$

At the alternative hypothesis, it is tested if it has increased, that is:

$H_1: \mu > 2.5$

We have the standard deviation for the population, thus, the z-distribution is used. The test statistic is given by:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

The parameters are:

$\overline{x}$ is the sample mean.
$\mu$ is the value tested at the null hypothesis.
$\sigma$ is the standard deviation of the sample.
n is the sample size.

For this problem, the values of the parameters are: $\overline{x} = 3, \mu = 2.5, \sigma = 1.5, n = 26$

Hence, the value of the test statistic is:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{3 - 2.5}{\frac{1.5}{\sqrt{26}}}$

$z = 1.7$

The critical value for a right-tailed test, as we are testing if the mean is greater than a value, with a significance level of 0.05, is of $z^{\ast} = 1.645$

Since the test statistic is greater than the critical value, it can be concluded that the mean length of jail time has increased.

A similar problem is given at brainly.com/question/24166849

5 0

2 years ago

You and three of your friends go out to dinner during the break.the check comes to 131.85$ you all decide to leave a 20% tip for

almond37 [142]

Answer:

$105.48

Step-by-step explanation:

Multiply 131.85 by 0.2 to get 20% of 131.85.

You will get 26.37.

Since it is 4 people, multiply by 4.

You will get $105.48.

4 0

2 years ago

Read 2 more answers

You bought a 5 pound turkey for $14.45. What is the cost for one pound?

sweet-ann [11.9K]

Answer:

$2.89

Step-by-step explanation:

$14.45/ 5= 2.89

3 0

2 years ago

4. Examine the graph of f(x) and the table that contains values of g(x). Which function has a greater rate of change over the in

Dominik [7]

g(x) has a greater rate of change in the interval 0<=x<=2.

The graph increases from -4 to -1, a change of +3

The table increases from 5 to 13, a change of +8.

5 0

3 years ago