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skelet666 [1.2K]

3 years ago

8

3) The average age of students at XYZ University is 24 years with a standard deviation of 8 years. Number of students at the uni

versity is 7500. A random sample of 36 students is selected. What is the probability that the sample mean will be between 25.5 and 27 years

1 answer:

otez555 [7]3 years ago

3 0

Answer:

0.1875

Step-by-step explanation:

σM=σ/√N

=8/√7500

=8/86.608

=0.092

Z=(x-μ)/σ/√N

=(25.5-36)/8/√7500

=-10.5/0.0092=-1141.304

Z score = -1.3125

=(27-36)/8/√7500 =

=9/0.0092=978.261

Z score= -1.125

-1.125-(-1.3125)=-1.125+1.3125)= 0.1875

The probability that the sample mean will be between 25.5 and 27 years

P(between 25.5 and 27) = 0.1875

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3 years ago

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The area of a circle is 3.142cm square.find the radius and diameter of the circle

suter [353]

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Read 2 more answers

How many randomly chosen guests should I invite to my party so that the probability of having a guest with the same birthday as

dimulka [17.4K]

Answer:

At least 401 chosen guests

Step-by-step explanation:

Represent those with same birth day with Same

So:

$P(Same) > \frac{2}{3}$

Represent number of people with n

There are 365 days in a year and 364 days out of these days is not your birthday.

So, there the probability that n people do not share your birthday is $(\frac{364}{365})^n$

i.e.

$P(none) = (\frac{364}{365})^n$

Solving further, we have that:

$P(same) + P(none) = 1$

$P(same) + (\frac{364}{365})^n = 1$

$P(same) = 1 - (\frac{364}{365})^n$

We're calculating the probability that $P(Same) > \frac{2}{3}$ .

So, we have:

$1 - (\frac{364}{365})^n > \frac{2}{3}$

Collect Like Terms

$1 - \frac{2}{3}> (\frac{364}{365})^n$

$\frac{3-2}{3}> (\frac{364}{365})^n$

$\frac{1}{3}> (\frac{364}{365})^n$

$3^{-1} > (\frac{364}{365})^n$

Take natural logarithm (ln) of both sides

$ln(3^{-1}) > ln((\frac{364}{365})^n)$

$-ln\ 3 > n\ ln\ (\frac{364}{365})$

Apply laws of logarithm

$-ln\ 3 > n\ (ln(364) - ln(365))$

Multiply through by -1

$ln\ 3 < n\ (ln(365) - ln(364))$

Solve for n

$\frac{ln\ 3}{(ln(365) - ln(364))} < n$

Reorder

$n > \frac{ln\ 3}{(ln(365) - ln(364))}$

$n > \frac{1.09861228867}{(5.89989735358 - 5.89715386764)}$

$n> \frac{1.09861228867}{0.00274348594}$

$n > 400.443928891$

$n > 400$ (approximated)

This implies that n = 401, 402, 403 .....

i.e

$n \geq 401$

So, at least 401 people has to be invited

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