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2 answers:

dangina [55]3 years ago

8 0

My name is no my sign is no my number is no you need to let it go

vlada-n [284]3 years ago

6 0

do u no de wea my bruddar

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The stright side of the hula hoop would be 2 inches

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4 years ago

Based on a survey conducted by Greenfield Online, 25 - 34-year-olds spend the most each week on fast food. The average weekly am

DIA [1.3K]

The correct answer is P = 0.1357.

In solving for the probability (P) that the 34-year olds will spend more than $60 on fastfood, the mean and the standard deviation is to be taken into account. The mean (μ) being the $44 average weekly expenditure; while the standard deviation (σ) is $14.50. The formula goes this way P(X > 60) = P( X−μ > 60−44 ) = P ( X−μ/σ > 60−44/14.50).

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3 years ago

Data were collected on the amount spent by 64 customers for lunch at a major Houston restaurant. These data are contained in the

galina1969 [7]

Answer:

a. The margin of error is 2.29.

b. 19.23 to 23.81

Step-by-step explanation:

The sample size is n=64.

We start by calculating the sample mean and standard deviation with the following formulas:

$M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}$

The sample mean is M=21.52.

The sample standard deviation is s=6.89.

We have to calculate a 99% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{6.89}{\sqrt{64}}=\dfrac{6.89}{8}=0.861$