2017 FALL MATH 2020 APPLIED STATISTICS (1) Nicole Hepch 10/12/17 2 48 PI Test: f

Question

2017 FALL MATH 2020 APPLIED STATISTICS (1) Nicole Hepch 10/12/17 2 48 PI Test: final exam first attempt Time 02:51:18 Sbmit This Question: 1 pt This Test: 30 pts possib Assume that d hat the method has no effect, so the different groups of couples use a particular method of gender selection and each e probabillty of a girl is 0.5. Assume that the groups consist of 23 couples Complete parts (a) through (c) below. coupie gives birth to one baby. This method is designed to increase the lkelhood that each baby will be a girt, but assume Find the mean and the standard deviation of the numbers of girls in groups of 23 biths. The value of the mean is (Type an integer or a decimal. Do not round.) The value of the standard deviation is (Round to one decimal place as needed.) b. Use the range rule of thumb to find the values separating results that are significantly low or signifcanty high Values ofgirts or fewer are signncantly low. Round to one decimal place as needed.) Wues of giris or geater are signicantiy high. (Round to one decimal place as needed.) 21gr. The result (Round to one decimal place as needed.)

Explanation / Answer

Ans:

Binomial distribution with n=23,p=0.5

a)mean=0.5*23=11.5

standard deviation=sqrt(23*0.5*(1-0.5))=sqrt(5.75)=2.398

b)7 or fewer girls are signifivantly low(as P(x<=7)<0.05

8 or greater are significantly high.

c)Result of 21 girls is not significantly high(as P(x=21)=0)

The method is not effective.

Use BINOMDIST function for the below table:

x P(x) P(X<=x) 0 0.0000 0.0000 1 0.0000 0.0000 2 0.0000 0.0000 3 0.0002 0.0002 4 0.0011 0.0013 5 0.0040 0.0053 6 0.0120 0.0173 7 0.0292 0.0466 8 0.0584 0.1050 9 0.0974 0.2024 10 0.1364 0.3388 11 0.1612 0.5000 12 0.1612 0.6612 13 0.1364 0.7976 14 0.0974 0.8950 15 0.0584 0.9534 16 0.0292 0.9827 17 0.0120 0.9947 18 0.0040 0.9987 19 0.0011 0.9998 20 0.0002 1.0000 21 0.0000 1.0000 22 0.0000 1.0000 23 0.0000 1.0000

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