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According to 2013 report from Population Reference Bureau, the mean travel time to work of workers ages 16 and older who did not

gregori [183]

Answer:

a) 48.80% probability that his travel time to work is less than 30 minutes

b) The mean is 30.7 minutes and the standard deviation is of 3.83 minutes.

c) 13.13% probability that in a random sample of 36 NJ workers commuting to work, the mean travel time to work is above 35 minutes

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 question, we have that:

$\mu = 30.7, \sigma = 23$

a. If a worker is selected at random, what is the probability that his travel time to work is less than 30 minutes?

This is the pvlaue of Z when X = 30. So

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

$Z = \frac{30 - 30.7}{23}$

$Z = -0.03$

$Z = -0.03$ has a pvalue of 0.4880.

48.80% probability that his travel time to work is less than 30 minutes

b. Specify the mean and the standard deviation of the sampling distribution of the sample means, for samples of size 36.

$n = 36$

Applying the Central Limit Theorem, the mean is 30.7 minutes and the standard deviation is $s = \frac{23}{\sqrt{36}} = 3.83$

c. What is the probability that in a random sample of 36 NJ workers commuting to work, the mean travel time to work is above 35 minutes?

This is 1 subtracted by the pvalue of Z when X = 35. So

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

By the Central Limit Theorem

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

$Z = \frac{35 - 30.7}{3.83}$

$Z = 1.12$

$Z = 1.12$ has a pvalue of 0.8687

1 - 0.8687 = 0.1313

13.13% probability that in a random sample of 36 NJ workers commuting to work, the mean travel time to work is above 35 minutes

8 0

4 years ago

Hey help me with ratios with double number lines uwur and you get 22 points

DENIUS [597]

Answer:

270 words in 3 minutes

Step-by-step explanation:

First find what the total on one minute is.

180/2 =90 words per minute.

So 180+90= 270 words in 3 minutes

7 0

3 years ago

Kayla is 13 years old or Uncle says that his h - 22 is equal to Kayla's age how old is Kayla's uncle

PolarNik [594]

H - 22 = 13

h = 35 years old Kayla's uncle

3 0

3 years ago

given that a1=5 and a2=15 are the first two terms of a geometric sequence, determine the values of a3 and a10. Show the calculat

Alekssandra [29.7K]

a_3 is 45 and a_10 is 98415

Step-by-step explanation:

Given

a_1 = 5

a_2 = 15

First of all we have to calculate the common ratio of the sequence. The common ratio is the ratio between two consecutive terms of a geometric sequence.

So,

$r=\frac{a_2}{a_1} = \frac{15}{5} = 3$