Answered

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A roadside assistance company gets millions of requests per year, with about a quarter of the requests in
summer and a quarter in winter. Suppose that 12% of all of the summer requests and 15% of the winter
requests are for help with a lockout. A data analyst pulls separate random samples of 200 summer and winter
requests and will look at the difference (summer winter) between the proportion of requests for help with
a lockout in each sample.
What are the mean and standard deviation of the sampling distribution of the difference in sample
proportions?

Sagot :

300 a summer plz I think

Answer: -0.03

sqrt ((0.12 (0.88) / 200) + (0.15)(0.85)/200)

Step-by-step explanation:

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