The student council is selling cupcakes at the school play.

Of 1,050 randomly selected adults, 360 identified themselves as manual laborers, 280 identified themselves as non-manual wage ea

garik1379 [7]

Answer:

We can claim with 95% confidence that the proportion of executives that prefer trucks is between 19.2% and 32.8%.

Step-by-step explanation:

We have a sample of executives, of size n=160, and the proportion that prefer trucks is 26%.

We have to calculate a 95% confidence interval for the proportion.

The sample proportion is p=0.26.

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.26*0.74}{160}}\\\\\\ \sigma_p=\sqrt{0.0012}=0.0347$

The critical z-value for a 95% confidence interval is z=1.96.

The margin of error (MOE) can be calculated as:

$MOE=z\cdot \sigma_p=1.96 \cdot 0.0347=0.068$

Then, the lower and upper bounds of the confidence interval are:

$LL=p-z \cdot \sigma_p = 0.26-0.068=0.192\\\\UL=p+z \cdot \sigma_p = 0.26+0.068=0.328$

The 95% confidence interval for the population proportion is (0.192, 0.328).

We can claim with 95% confidence that the proportion of executives that prefer trucks is between 19.2% and 32.8%.

3 0

2 years ago

En una cancha se va a pintar la circunferencia central que tiene un diámetro de 5 m .¿cual es la longitud de la circunferencia?

Alona [7]

Answer:

The length of the circumference is $5\pi\ m$ or $15.70\ m$

Step-by-step explanation:

The question in English is

On a court, the central circumference will be painted, which has a diameter of 5 m. What is the length of the circumference?

we know that

The circumference of a circle is given by the formula

$C=\pi D$

we have

$D=5\ m$

substitute

$C=\pi (5)\\C=5\pi\ m$

This is the exact value

assume

$\pi=3.24$

$C=5(3.14)=15.70\ m$

4 0

1 year ago

Victor fue al mercado para comprar manzanas, naranjas y platanos; las naranjas costaron el doble de lo 1ue pago por las manzanas

Thepotemich [5.8K]

Answer:

El precio de las manzanas = 27 pesos

El precio de las naranjas = 54 pesos

El precio de las bananas = 19 pesos

Step-by-step explanation:

Los parámetros dados son;

El monto total gastado = 100 pesos

Sea el precio de las naranjas = x

Sea el precio de las manzanas = y

Sea el precio de los plátanos = z

La cantidad pagada por las naranjas = 2 · y = x

La cantidad pagada por los plátanos = y - 8 = z

Por lo tanto, tenemos;

La cantidad total gastada = La cantidad pagada por las naranjas + La cantidad pagada por las bananas + La cantidad pagada por las manzanas

∴ El monto total gastado = 100 pesos = 2 · y + y - 8 + y

100 = 4 · años - 8

4 · y = 100 + 8 = 108

y = 108/4 = 27

y = 27

De

z = y - 8 tenemos;

z = 27 - 8 = 19

De 2 · y = x, tenemos;

2 × 27 = x

x = 54

Por lo tanto;

El precio de las naranjas = 54 pesos

El precio de las manzanas = 27 pesos

El precio de los plátanos = 19 pesos.

5 0

1 year ago

For a normally distributed random variable x with m = 75 and s = 4, find the probability that 69 < x < 79 Use the table to

nalin [4]

10-14-19-12-14-18-10-15-15

6 0

2 years ago

The average annual amount American households spend for daily transportation is $6312 (Money, August 2001). Assume that the amou

lions [1.4K]

Answer:

(a) The standard deviation of the amount spent is $3229.18.

(b) The probability that a household spends between $4000 and $6000 is 0.2283.

(c) The range of spending for 3% of households with the highest daily transportation cost is $12382.86 or more.

Step-by-step explanation:

We are given that the average annual amount American households spend on daily transportation is $6312 (Money, August 2001). Assume that the amount spent is normally distributed.

(a) It is stated that 5% of American households spend less than $1000 for daily transportation.

Let X = the amount spent on daily transportation

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = average annual amount American households spend on daily transportation = $6,312

$\sigma$ = standard deviation

Now, 5% of American households spend less than $1000 on daily transportation means that;

P(X < $1,000) = 0.05

P( $\frac{X-\mu}{\sigma}$ < $\frac{\$1000-\$6312}{\sigma}$ ) = 0.05

P(Z < $\frac{\$1000-\$6312}{\sigma}$ ) = 0.05

In the z-table, the critical value of z which represents the area of below 5% is given as -1.645, this means;

$\frac{\$1000-\$6312}{\sigma}=-1.645$

$\sigma=\frac{-\$5312}{-1.645}$ = 3229.18

So, the standard deviation of the amount spent is $3229.18.

(b) The probability that a household spends between $4000 and $6000 is given by = P($4000 < X < $6000)

P($4000 < X < $6000) = P(X < $6000) - P(X $\leq$ $4000)

P(X < $6000) = P( $\frac{X-\mu}{\sigma}$ < $\frac{\$6000-\$6312}{\$3229.18}$ ) = P(Z < -0.09) = 1 - P(Z $\leq$ 0.09)

= 1 - 0.5359 = 0.4641

P(X $\leq$ $4000) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{\$4000-\$6312}{\$3229.18}$ ) = P(Z $\leq$ -0.72) = 1 - P(Z < 0.72)

= 1 - 0.7642 = 0.2358

Therefore, P($4000 < X < $6000) = 0.4641 - 0.2358 = 0.2283.

(c) The range of spending for 3% of households with the highest daily transportation cost is given by;

P(X > x) = 0.03 {where x is the required range}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-\$6312}{3229.18}$ ) = 0.03

P(Z > $\frac{x-\$6312}{3229.18}$ ) = 0.03

In the z-table, the critical value of z which represents the area of top 3% is given as 1.88, this means;

$\frac{x-\$6312}{3229.18}=1.88$

${x-\$6312}=1.88\times 3229.18$

x = $6312 + 6070.86 = $12382.86

So, the range of spending for 3% of households with the highest daily transportation cost is $12382.86 or more.

8 0

2 years ago