Research in March 2019 suggests that 40% of U. S. adults approve of way President Trump is running the country. We randomly sample 50 U. S. adults and find that 35% approve of way President Trump is running the country. What is the probability that a random sample of 50 U. S. adults has less than 35% with this opinion

0.235

Step-by-step explanation:

Define S as

S = the event that a randomly selected adult in the U.S. approves of President Trump.

Define ~S as the negation of S, i.e.,

~S = the event that a randomly selected adult in the U.S. does not approve of President Trump

Then,

P(S) = 40% = 0.4

P(~S) = 1 - 0.4 = 0.6

35% of 50 = 0.35 * 50 = 17.5.

We want to compute:

P(a random selection of 50 adults has less than 17.5 of them approving of President Trump) =

P(17 of the 50 approve) + P(16 of the 50 approve) + P(15 of the 50 approve) + . . . + P(none of the 50 approve).  

We approximately compute this by approximating the summation (which is from a binomial distribution) by a normal distribution with

mean = 0.4(50) = 20,

and

standard deviation = sqrt( .4 * .6 * 50) = sqrt( 12 ) = 3.464 rounded to the nearest 1000th.  

So, for normal random variable X with mean = 20 and std = 3.464, the probability that we want is approximately equal to P(X < 17.5), where we note that X is a real random variable.  

So, the Z value = (17.5 - 20)/3.464 = -0.722, rounded to the nearest 1000th.  

Using the normal random variable online calculator, https://stattrek.com/online-calculator/normal.aspx,

we get that

P(z < -0.722) = 0.235, which is the answer to the question.  

it would open into a square depending on the type of envelope i’m not sure about the second part