I NEED HELP FAST!!!

How much time do Americans spend eating or drinking? Suppose for a random sample of 1001 Americans age 15 or older, the mean amo

levacccp [35]

Answer:

Step-by-step explanation:

a)

To construct a confidence interval ,the histogram should be bell shaped.It means that the it should be normally distributed with no skewness.To eliminate the skewness a large sample size is required so that the sample is normally distribute about the mean Also ,in order to construct a t-interval the sample data must come from the population that is normally distributed or the sample size is larger than 30.Since the question stated that the population distribution is skewed to the right \bar{x} is guaranteed to be normally distributed if $n\geq 30$

b)The sample satisfies the normal distribution because the sample sixe is greater than 30 which is 1001.

c) [ $\bar{x}=1.22 \,s=.65\, n=1001$

Area in the right tail =2.5% or .025

degree of freedom =1001-1 =1000

$t_{\alpha /2}=1.96$

95% confidence interval is : $\bar{x}\pm t_{\alpha /2}s/\sqrt{n}
$

lower bound is : $1.22-1.96*.65/\sqrt{1001} =1.18
$

upper bound is : $-1.22+1.96*.65/\sqrt{1001}=1.26
$

d)No because the sample data was obtined from Americans age 15 or older which cannot be applied to a different age group

8 0

3 years ago

Subtract: 9-7 Subtracting 9-7=2

Licemer1 [7]

Answer: 2

Step-by-step explanation: add 2 to 7 you get 9

6 0

3 years ago

A company services home air conditioners. It is known that times for service calls follow a normal distribution with a mean of 7

SCORPION-xisa [38]

Answer:

The probability that exactly eight of them take more than 93.6 minutes is 5.6015 $\times 10^{-6}$ .

Step-by-step explanation:

We are given that it is known that times for service calls follow a normal distribution with a mean of 75 minutes and a standard deviation of 15 minutes.

A random sample of twelve service calls is taken.

So, firstly we will find the probability that service calls take more than 93.6 minutes.

Let X = times for service calls.

So, X ~ Normal( $\mu=75,\sigma^{2} =15^{2}$ )