# Help with statistics homework

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In San Francisco 30% of workers take public transportation daily (usa today December 21 2005

a) in a sample of 10 workers what is the probability that exactly three workers take public transportation daily (4 decimals including interim calculations

b) in a sample of 10 workers what is the probability that at least three workers take public transportation daily (4 decimals including interim calculations

Oct 20th, 2015

Thank you for the opportunity to help you with your question!

1. P[x = 3] = 10C3 * 0.3^3 * 0.7^3
= 0.1852
= 18.52%

2. P[x >= 3] = P[x = 3] + P[x = 4] + P[x = 5] + P[x = 6] + P[x=7] + P[x=8] + P[x=9] + P[x=10]

P[x = 3] = 10C3 * 0.3^3 * 0.7^3 = 0.1852
P[x = 4] = 10C4 * 0.3^4 * 0.7^2 = 0.0595
P[x = 5] = 10C5 * 0.3^5 * 0.7^1 = 0.0102
P[x = 6] = 10C6 * 0.3^6 * 0.7^0 = 0.0007

P[x=7]=10C7 * 0.3^7 0.7^0=

P[x=8]=10C8 * 0.3^8 * 0.7^8=

P[x=9]=10C9 * 0.3^9 * 0.7^1=

P[x=10]=10C10 * 0.3^10 * 0.7^0=

P[x >= 3] = 0.1852 + 0.0595 + 0.0102 + 0.0007
=

Please let me know if you need any clarification. I'm always happy to answer your questions.
Oct 20th, 2015

don't withdraw please!  answer following soon...20 minutes ran out...please don't withdraw i'm finishing ASAP!

Oct 20th, 2015

don't withdraw please!  answer following soon...20 minutes ran out...please don't withdraw i'm submitting ASAP!

Oct 20th, 2015

This is binomial distribution, I need some more minutes to grasp something...Please dear don't withdraw (it's extremely consequential to me)

Oct 20th, 2015

1. P[x = 3] = 10C3 * 0.3^3 * 0.7^3
= 0.2668
= 26.68%

2. P[x >= 3] = P[x = 3] + P[x = 4] + P[x = 5] + P[x = 6] +P[x=7] +P[x=8] +P[x=9] + P[x=10]

Probability that at least three take public transport daily=0.0028

Oct 20th, 2015

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Oct 20th, 2015
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Oct 20th, 2015
Nov 25th, 2017
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