Which of the following functions computes a value such that 2.5% of the area under the standard normal distribution lies in the upper tail defined by this value? ○ 0 ° ○ a.-normsinv(0.975) b.-norm·s. inv(0.05) c.-norm. s-inv(0.95) d.-normsinv(0.025)

a) NORM.S.INV(0.975)

Step-by-step explanation:

1) Some definitions

The standard normal distribution is a particular case of the normal distribution. The parameters for this distribution are: the mean is zero and the standard deviation of one. The random variable for this distribution is called Z score or Z value.

NORM.S.INV Excel function "is used to find out or to calculate the inverse normal cumulative distribution for a given probability value"

The function returns the inverse of the standard normal cumulative distribution(a z value). Since uses the normal standard distribution by default the mean is zero and the standard deviation is one.

2) Solution for the problem

Based on this definition and analyzing the question :"Which of the following functions computes a value such that 2.5% of the area under the standard normal distribution lies in the upper tail defined by this value?".

We are looking for a Z value that accumulates 0.975 or 0.975% of the area on the left and by properties since the total area below the curve of any probability distribution is 1, then the area to the right of this value would be 0.025 or 2.5%.

So for this case the correct function to use is: NORM.S.INV(0.975)

And the result after use this function is 1.96. And we can check the answer if we look the picture attached.

42%

step-by-step explanation:

we know,

total people = 60

total males = 25

percentage of males

= 25/60 × 100

= 25/6 × 10

= 25/3 × 5

= 125 / 3

= 41.

rounding of = 42%

happy to

pls mark as brainliest

your answer would be 12