jrrainbow11
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1.An electronic prize wheel is programmed to land on "win” 25% of the time on weekdays and 10% of the
time on weekends, and consecutive plays produce winners independently. Suppose we observe a random
sample of 50 outcomes from this machine on a weekday and 100 outcomes from this machine on a
weekend. Which of the following descritts the shape, mean, and standard deviation of the sampling
distribution of the difference in the two sample proportions?
(a) Shape = approximately normal, mean = 0.15, standard deviation = 0.07
(b) Shape = approximately normal, mean = 0.15, standard deviation = 0.05
(c) Shape = approximately normal, mean = 0.35, standard deviation = 0.07
(d) Shape = skewed right, mean = 0.35, standard deviation = 0.05
(e) Shape = skewed right, mean 0.35, standard deviation = 0.0

Sagot :

janthonymorales

Answer: D

Step-by-step explanation:

The shape, mean, and standard deviation of the sampling distribution of the difference in the two sample proportions is Shape = skewed right, mean = 0.35, standard deviation = 0.05.

What is the standard deviation?

In statistics, the standard deviation of a random variable, data is the square root of its variance.

An electronic prize wheel is programmed to land on "win” 25% of the time on weekdays and 10% of the time on weekends, and consecutive plays produce winners independently.

A random sample of 50 outcomes from this machine on a weekday and 100 outcomes from this machine on a weekend.

[tex]P_{1} = 25%\\ =0.25\\P_{2} = 0.10\\[/tex]

[tex]n_{1} = 50\\n_{2} =100[/tex]

Let x be a random variable for weekdays

[tex]x= N[0.25, 0.25\times(1-0.25)/50]\\x = N[0.25, 0.00375][/tex]

Let y be randomly variable for weekdays

[tex]Y= N[0.10, 0.10\times(1-0.10)/100]\\Y= N[0.10, 0.0009][/tex]

Thus, the shape, mean, and standard deviation of the sampling distribution of the difference in the two sample proportions is Shape = skewed right, mean = 0.35, standard deviation = 0.05.

Learn more about standard deviation;

https://brainly.com/question/11078015

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