Help ASAP plzzzz!!!! T^T

A study conducted by the Center for Population Economics at the University of Chicago studied the birth weights of 614 babies bo

MissTica

Answer:

The number of newborns who weighed between 1614 grams and 5182 grams was of 586.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The mean weight was 3398 grams with a standard deviation of 892 grams.

This means that $\mu = 3398, \sigma = 892$

Proportion that weighed between 1614 and 5182 grams:

p-value of Z when X = 5182 subtracted by the p-value of Z when X = 1614.

X = 5182

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5182 - 3398}{892}$

$Z = 2$

$Z = 2$ has a p-value of 0.9772

X = 1614

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1614 - 3398}{892}$

$Z = -2$

$Z = -2$ has a p-value of 0.0228

0.9772 - 0.0228 = 0.9544.

Out of 614 babies:

0.9544*614 = 586

The number of newborns who weighed between 1614 grams and 5182 grams was of 586.

4 0

3 years ago

(1 point) A recent report for a regional airline reported that the mean number of hours of flying time for its pilots is 56 hour

Ray Of Light [21]

Answer:

(53.812 ; 58.188) ; 156

Step-by-step explanation:

Given that :

Sample size (n) = 51

Mean (m) = 56

Standard deviation (σ) = 9.5

α = 90%

Using the relation :

Confidence interval = mean ± Error

Error = Zcritical * (standard deviation / sqrt (n))

Zcritical at 90% = 1.645

Error = 1.645 * (9.5 / sqrt(51))

Error = 1.645 * 1.3302660

Error = 2.1882877

Hence,

Confidence interval :

Lower boundary = 56 - 2.1882877 = 53.8117123

Upper boundary = 56 + 2.1882877 = 58.1882877

Confidence interval = (53.812 ; 58.188)

2.)

Margin of Error (ME) = 1.25

α = 90%

Sample size = ((Zcritical * σ) / ME)^2

Zcritical at 90% = 1.645

Sample size = ((1.645 * 9.5) / 1.25)^2

Sample size = (15.6275 / 1.25)^2

Sample size = 12.502^2 = 156.3000

Sample size = 156

3 0

3 years ago

Need answers quick its due in 1 hour!!!!!

antiseptic1488 [7]

Answer:

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i made a video explaining its on

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

A marketing firm would like to test-market the name of a new energy drink targeted at 18- to 29-year-olds via social media. A st

Anon25 [30]

Answer:

(a) The probability that a randomly selected U.S. adult uses social media is 0.35.

(b) The probability that a randomly selected U.S. adult is aged 18–29 is 0.22.

(c) The probability that a randomly selected U.S. adult is 18–29 and a user of social media is 0.198.

Step-by-step explanation:

Denote the events as follows:

X = an US adult who does not uses social media.

Y = an US adult between the ages 18 and 29.

Z = an US adult between the ages 30 and above.

The information provided is:

P (X) = 0.35

P (Z) = 0.78

P (Y ∪ X') = 0.672

(a)

Compute the probability that a randomly selected U.S. adult uses social media as follows:

P (US adult uses social media (X')) = 1 - P (US adult so not use social media)

$=1-P(X)\\=1-0.35\\=0.65$

Thus, the probability that a randomly selected U.S. adult uses social media is 0.35.

(b)

Compute the probability that a randomly selected U.S. adult is aged 18–29 as follows:

P (Adults between 18 - 29 (Y)) = 1 - P (Adults 30 or above)

$=1-P(Z)\\=1-0.78\\=0.22$

Thus, the probability that a randomly selected U.S. adult is aged 18–29 is 0.22.

(c)

Compute the probability that a randomly selected U.S. adult is 18–29 and a user of social media as follows:

P (Y ∩ X') = P (Y) + P (X') - P (Y ∪ X')

$=0.22+0.65-0.672\\=0.198$

Thus, the probability that a randomly selected U.S. adult is 18–29 and a user of social media is 0.198.

6 0

3 years ago

Help me with this please!

DedPeter [7]

Answer:

Area of triangle = 20√2

Step-by-step explanation:

Formula:

Area of triangle = bh/2

b - base of triangle

h - height of triangle

Distance formula

d = √(x₂ - x₁)² +(y₂ - y₁)²

From the figure we can see that, the given triangle is right angled triangle.

Base = AC and Height = AB

To find AB and AC

A(1,3), B(4,7) and C(9, -5)

AB = √(4 - 1)² +(7 - 3)² = √(3² + 4²) = 5

AC = √(9 - 1)² +(-5 - 3)² = √(8² + -8²) = 8√2

To find the are of triangle

Area = bh/2 = (5* 8√2)/2 = 20√2

Therefore the area of triangle = 20√2 square units

8 0

3 years ago

Read 2 more answers