1. Let's assign numerical values to coin flips: Heads is +1 and Tails is –1, each with probability ½. After a huge number of coin flips, you'd expect an average value of µ = 0, because you'd expect roughly as many Heads (+1) as Tails (-1), so on average they would cancel out to 0. Since Heads and Tails are both 1 unit of deviation away from the mean, the standard deviation is just o = 1. Suppose that you flip a coin n = 100 times. a) If you get 49 Heads and 51 Tails, what is the “mean coin flip score" x? Hint: just apply the usual definition of mean: add up all the values, with +1 for each Head and -1 for each Tail, then divide by 100. b) Spoiler alert: the answer to part (a) is = -0.02. Use this to answer the following: what is the probability that, if you flip a coin 100 times, you get 49 Heads or fewer? Hint: calculate the z-score for this sample, using z = (x - µ) / (0/Vn), and then use the chart. c) Now repeat the procedure to figure out the probability that, if you flip a coin 100 times, you get 45 Heads or fewer. d) Finally, what are the odds of getting just 40 Heads or fewer out of 100 coin flips?

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hello can you please answer 1a 1b 1c and i know you can’t answer 1d but if you could that would be very appreciated thank you so much!
1. Let's assign numerical values to coin flips: Heads is +1 and Tails is –1, each with probability ½.
After a huge number of coin flips, you'd expect an average value of µ = 0, because you'd expect roughly
as many Heads (+1) as Tails (-1), so on average they would cancel out to 0. Since Heads and Tails are
both 1 unit of deviation away from the mean, the standard deviation is just o = 1.
Suppose that you flip a coin n = 100 times.
a) If you get 49 Heads and 51 Tails, what is the “mean coin flip score" x? Hint: just apply the usual
definition of mean: add up all the values, with +1 for each Head and –1 for each Tail, then divide by 100.
b) Spoiler alert: the answer to part (a) is š = -0.02. Use this to answer the following: what is the
probability that, if you flip a coin 100 times, you get 49 Heads or fewer? Hint: calculate the z-score for
this sample, using z =
(x – µ) / (0/Vn), and then use the chart.
c) Now repeat the procedure to figure out the probability that, if you flip a coin 100 times, you get 45
Heads or fewer.
d) Finally, what are the odds of getting just 40 Heads or fewer out of 100 coin flips?
Transcribed Image Text:1. Let's assign numerical values to coin flips: Heads is +1 and Tails is –1, each with probability ½. After a huge number of coin flips, you'd expect an average value of µ = 0, because you'd expect roughly as many Heads (+1) as Tails (-1), so on average they would cancel out to 0. Since Heads and Tails are both 1 unit of deviation away from the mean, the standard deviation is just o = 1. Suppose that you flip a coin n = 100 times. a) If you get 49 Heads and 51 Tails, what is the “mean coin flip score" x? Hint: just apply the usual definition of mean: add up all the values, with +1 for each Head and –1 for each Tail, then divide by 100. b) Spoiler alert: the answer to part (a) is š = -0.02. Use this to answer the following: what is the probability that, if you flip a coin 100 times, you get 49 Heads or fewer? Hint: calculate the z-score for this sample, using z = (x – µ) / (0/Vn), and then use the chart. c) Now repeat the procedure to figure out the probability that, if you flip a coin 100 times, you get 45 Heads or fewer. d) Finally, what are the odds of getting just 40 Heads or fewer out of 100 coin flips?
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