The times it took for 35 loggerhead sea turtle eggs to hatch in a simple random sample are normally distributed, with a mean of 50 days and a standard deviation of 2 days. assuming a 95% confidence level (95% confidence level = z-score of 1.96), what is the margin of error for the population mean? remember, the margin of error, me, can be determined using the formula . 0.06 0.11 0.34 0.66

Answer-

The margin of error is 0.66

Solution-

The times it took for 35 loggerhead sea turtle eggs to hatch in a simple random sample are normally distributed, with a mean of 50 days and a standard deviation of 2 days.

So here,

sample size = n = 35

confidence level (95%) = z = 1.96

mean = μ = 50

standard deviation = s = 2

We know that,

Putting the values,

Therefore, the margin of error is 0.66

answer: 4/6 =5/7.5

step-by-step explanation: it is based on scale and the scale between the two is 1.25. the answer is correct but i dont know if i explained it right

true? is that what you're asking or do you mean something else?