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asambeis [7]
3 years ago
9

In a popular online role playing game, players can create detailed designs for their character's "costumes," or appearance. Mica

ela sets up a website where players can buy and sell these costumes online. Information about the number of people who visited the website and the number of costumes purchased in a single day is listed below.
196 visitors purchased no costume.
108 visitors purchased exactly one costume.
22 visitors purchased more than one costume.

Based on these results, express the probability that the next person will purchase more than one costume as a percent to the nearest whole number.
Mathematics
1 answer:
Veronika [31]3 years ago
5 0

Answer:     6.75% which rounds up to 7%

Step-by-step explanation:

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Suppose that the mean water hardness of lakes in Kansas is 425 mg/L and these values tend to follow a normal distribution. A lim
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Answer:

The mean water hardness of lakes in Kansas is 425 mg/L or greater.

Step-by-step explanation:

We are given the following data set:

346, 496, 352, 378, 315, 420, 485, 446, 479, 422, 494, 289, 436, 516, 615, 491, 360, 385, 500, 558, 381, 303, 434, 562, 496

Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{10959}{25} =438.36

Sum of squares of differences = 175413.76

S.D = \sqrt{\dfrac{175413.76}{24}} = 85.49

Population mean, μ = 425 mg/L

Sample mean, \bar{x} = 438.36

Sample size, n = 25

Alpha, α = 0.05

Sample standard deviation, s = 85.49

First, we design the null and the alternate hypothesis

H_{0}: \mu \geq 425\text{ mg per Litre}\\H_A: \mu < 425\text{ mg per Litre}

We use one-tailed t test to perform this hypothesis.

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{438.36 - 425}{\frac{85.49}{\sqrt{25}} } = 0.7813

Now, t_{critical} \text{ at 0.05 level of significance, 24 degree of freedom } = -1.7108

Since,                        

The calculated t-statistic is greater than the critical value, we fail to reject the null hypothesis and accept it.

Thus, the mean water hardness of lakes in Kansas is 425 mg/L or greater.

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