In a recent poll, 44% 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 150 randomly selected students on your campus and find that 71 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 71 would prefer a boy, assuming the true percentage is 44%.
(b) Does this result contradict the poll? Explain.

a) The probability that 71 of 150 will prefer boy child is 71/150 or 0.47

b) The result contradicts the poll actual percentage is 47.33% which is 3.33% more than the poll predicted

Step-by-step explanation:

If 71 out of 150 prefer boy child

The probability that the 71 will prefer boy child is

= 71/150

The actual percentage is

(71/150)*100%

= 47.33%

This contradicts the poll as this is more than the poll predicted. That means Less than 71 of 150 actually preferred boy child.

333.74 ! should be your answer for this question

0 and 5 minutes

step-by-step explanation:

given that one morning eight buses arrive late by the number of minutes given below:

0 8 2 6 9 2 0 7 8

write these in ascending order

0   0 2 6 7 8 8 9

in the afternoon two more buses arrive at the bus stop say by x and y minutes late.

total number of buses = 10

median = average of 4th and 5th after arranging in ascending order

since 4th entry is 6, the values of x and y should be below 6

also given 0 is the mode so either x or y or both should be 0 because then only 0 gets the maximum frequency and hence mode

thus we have

0 0 0 2 6 7 8 8 9 and another unknown entry which is below 6.

since median is 3.5, we have the unknown entry to be 5 only because then only median =( 2+5)/2

thus the after noon buses were late by 0 and 5 minutes.