In a recent poll, 35% of survey respondents said that, if they only had one child, they would prefer the child to be a boy. Suppose you conducted a survey of 160 randomly selected students on your campus and find that 73 of them would prefer a boy. a. Use the normal approximation to the binomial to approximate the probability that, in a random sample of 150 students, at least 74 would prefer a boy, assuming the true percentage is 38%. b. Does this result contradict the poll? Explain.

we have that z = 1.8326 which is not close to z factor of 0. The result is not within limits.

Step-by-step explanation:

Following normal approximation to binomial we have

mean = n×p

standard deviation = √(n×p×(1-p))

n is equal to 160

p is equal to 35%

n×p is equal to 150×0.35 = 52.5

this is the probability that you would expect based on the knowledge that 35% of survey respondents said they would prefer a boy than a girl.

in your survey you got 74 out of 150 and you want to know if this is statistically significant or if you can expect this deviation by chance.

standard deviation is calculated as √(n×p× (1-p))which becomes

√ (150×0.35×0.75 = 6.2749 which is more than enough accuracy for what is required.

without a continuity correction, your formula would be:

z (74 > 52.5) = (population mean - sample mean) / population standard deviation which becomes (74-52.5)/6.2749 = 1.8326

this is not close to a z factor of 0 which means that the results are not within limits.

with a continuity correction, your formula would be:

z (52.5 > 52.5) = 0

A z factor of zero means that your sample mean is exactly the population mean.

translate k to n and reflect across the line containing jk.

i need points

step-by-step explanation:

sorry bubs