# According to a study, the probability that a randomly selected teenagar shopped at a mall at least once during a week was 0.61.

###### Question:

## Answers

Step-by-step explanation:

Given that:

p = 0.61

If X is the the number of students in a randomly selected group of a sample size n = 50

The expected value and the standard deviation can be computed as follows:

The expected value E(X) = np

The expected value E(X) = 50 × 0.61

The expected value E(X) = 30.5

The required standard deviation = [tex]\sqrt{np(1-p)}[/tex]

The required standard deviation = [tex]\sqrt{30.5(1-0.61)}[/tex]

The required standard deviation = [tex]\sqrt{30.5(0.39)}[/tex]

The required standard deviation = [tex]\sqrt{11.895}[/tex]

The required standard deviation = 3.4489

The required standard deviation = 3.45

(b) Fill in the missing quantity. (Round your answer to the nearest whole number.)

There is an approximately 2.5% chance that _____ or more teenagers in the group will shop at the mall during the next week.

From the given information:

Now, we can deduce that:

the mean = 30.5

standard deviation = 3.45

Using the empirical rule:

At 95% confidence interval;

[μ - 2σ, μ + 2σ] = [ 30.5 - 2(3.45) , 30.5 + 2(3.45)]

[μ - 2σ, μ + 2σ] = [ 30.5 - 6.9 , 30.5 + 6.9]

[μ - 2σ, μ + 2σ] = [ 23.6, 37.4]

The 2.5% of the observations are less than 95% confidence interval and 2.5% observations are greater than 95% confidence interval.

The required number of teenagers is = the upper limit of the 95% confidence interval = 37

There is an approximately 2.5% chance that __37___ or more teenagers in the group will shop at the mall during the next week.

6.

Cos60 = x/16

x = 16cos60

x = 16*√3/2

x = 8√3

7)

a = √(8.5² - 4.2²)

a ≈ 7.39 ≈ 8cm

8)

a² = 165² + 220²

a = √(165² + 220²)

a = 275

90 is the correct answer

1 hour 20 minutes

Step-by-step explanation:

Part A:

Maria

1/4 x x/4

Josie

1/2 x x/2

Part B:

1/4x + 1/2x

= 1/x

3/4x = 1/x

x = 4/3 hr

= 1 hour 20 minutes

Together they can complete job in 1 hour and 20 minutes

I'm sorry I can't help you with Q3 because there's no table for that problem.

Still hope this helps!

1 hr 20 min

Step-by-step explanation:

Part A

Maria

1/4 x x/4

Josie

1/2 x x/2

Part B

1/4x+1/2x= 1/x

3/4x=1/x

x= 4/3 hr = 1 hr 20 min

Together they can complete job in 1 hr and 20 min

Question 3

missing table

1

Step-by-step explanation:

cos^-1(0.39)=1.17... so i rounded to the nearest whole number, 1

67°

Step-by-step explanation:

Apply the inverse cosine function to both sides, obtaining:

x = arccos 0.39

Using a calculator, obtain the value of x:

x = 1.17 (radians)

or

1.17 180°

x = · = 67 degrees (to the nearest whole degree)

1 π

1

a) [tex]E(X) = np=50*0.67=33.50[/tex]

[tex]sd(X)=\sqrt{np(1-p)}=\sqrt{50*0.67*(1-0.67)}=3.32[/tex]

b) There is an approximately 2.5% chance that 27 or more teenagers in the group will shop at the mall during the next week.

Step-by-step explanation:

1) Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

2) Part a

Let X the random variable of interest, on this case we now that:

[tex]X \sim Binom(n=50, p=0.67)[/tex]

The probability mass function for the Binomial distribution is given as:

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]

Where (nCx) means combinatory and it's given by this formula:

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]

The expected value is given by this formula:

[tex]E(X) = np=50*0.67=33.50[/tex]

And the standard deviation for the random variable is given by:

[tex]sd(X)=\sqrt{np(1-p)}=\sqrt{50*0.67*(1-0.67)}=3.32[/tex]

3) Part b

We need to check if we can use the normal approximation , the conditions are:

[tex]np=50*0.67=33.5010[/tex] and [tex]n(1-p)=50*(1-0.67)=16.510[/tex]

So then we can apply the normal approximation to the binomial distribution in our case:

[tex]X \sim N(\mu=33.50,\sigma=3.32)[/tex]

We need on the right tail of the distribution a value a that gives to us 2.5% of the area below and 97.5% of the area above. Both conditions are equivalent.

Let's use the condition [tex]P(X<a)=0.025[/tex], the best way to solve this problem is using the z score with the following formula:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

So we need a value from the normal standard distribution that accumulates 0.025 of the area on the left and 0.975 on the right. This value on this case is -1.96 and we can founded with the following code in excel:

"=NORM.INV(0.025,0,1)"

If we apply the z score formula to our case we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-33.5}{3.32})=P(Z<-1.96)=0.025[/tex]

So then based on the equalities we have this:

[tex]\frac{a-33.5}{3.32}=-1.96[/tex]

And if we solve for a we got:

[tex]a=(-3.32*1.96) +33.50=26.99[/tex]

There is an approximately 2.5% chance that 27 or more teenagers in the group will shop at the mall during the next week.

We have to find the missing value to the nearest whole number of tan x° =0.9.

Since, tan x° =0.9

To evaluate the missing value of 'x', we will take [tex]\arctan (0.9)[/tex]

So, [tex]x^{\circ}=\arctan (0.9)[/tex]

Now, we will find the value of [tex]\arctan (0.9)[/tex]using the calculator, we get,

[tex]x^{\circ}=\arctan (0.9)=41.9^{\circ}=42^{\circ}[/tex]

So, the missing value of 'x' to the nearest whole number is [tex]42^{\circ}[/tex].