Find the area under the standard normal curve to the left of z = 1.5

(40pts) Mr. Green had 200 dollars and Mrs.Green had 180 dollars. After they each bought the same t-shirt, the ratio of the numbe

yanalaym [24]

Answer:

The t-shirt cost $23.33.

Step-by-step explanation:

6 0

2 years ago

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A study of immunizations among school‑age children in California found that some areas had rates of unvaccinated school‑age chil

Rom4ik [11]

Answer:

Probability that none of the 20 children in such a classroom would be unvaccinated is 0.055.

Step-by-step explanation:

We are given that a classroom of 20 children in one such area where 13.5% of children are unvaccinated.

If there are no siblings in the classroom, we are willing to consider the vaccination status of the 2020 unrelated children to be independent.

The above situation can be represented through binomial distribution;

$P(X=r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r} ; x = 0,1,2,3,......$

where, n = number of trials (samples) taken = 20 children

r = number of success = none of the 20 children

p = probability of success which in our case is probability that

children are unvaccinated, i.e; p = 13.5%

Let X = Number of children that are unvaccinated

So, X ~ Binom(n = 20, p = 0.135)

Now, Probability that none of the 20 children in such a classroom would be unvaccinated is given by = P(X = 0)

P(X = 0) = $\binom{20}{0} \times 0.135^{0} \times (1-0.135)^{20-0}$

= $1\times 1 \times 0.865^{20}$

= 0.055

Hence, the probability that none of the 20 children in such a classroom would be unvaccinated is 0.055.

8 0

2 years ago

A Roper survey reported that 65 out of 500 women ages 18-29 said that they had the most say when purchasing a computer; a sample

8090 [49]

Answer:

Step-by-step explanation:

Step(i):-

Given first random sample size n₁ = 500

Given Roper survey reported that 65 out of 500 women ages 18-29 said that they had the most say when purchasing a computer.

First sample proportion

$p^{-} _{1} = \frac{65}{500} = 0.13$

Given second sample size n₂ = 700

Given a sample of 700 men (unrelated to the women) ages 18-29 found that 133 men said that they had the most say when purchasing a computer.

second sample proportion

$p^{-} _{2} = \frac{133}{700} = 0.19$

Level of significance = α = 0.05

critical value = 1.96

Step(ii):-

Null hypothesis : H₀: There is no significance difference between these proportions

Alternative Hypothesis :H₁: There is significance difference between these proportions

Test statistic

$Z = \frac{p_{1} ^{-}-p^{-} _{2} }{\sqrt{PQ(\frac{1}{n_{1} } +\frac{1}{n_{2} } )} }$

where

$P = \frac{n_{1} p^{-} _{1}+n_{2} p^{-} _{2} }{n_{1}+ n_{2} } = \frac{500 X 0.13+700 X0.19 }{500 + 700 } = 0.165$

Q = 1 - P = 1 - 0.165 = 0.835

$Z = \frac{0.13-0.19 }{\sqrt{0.165 X0.835(\frac{1}{500 } +\frac{1}{700 } )} }$

Z = -2.76

|Z| = |-2.76| = 2.76 > 1.96 at 0.05 level of significance

Null hypothesis is rejected at 0.05 level of significance

Alternative hypothesis is accepted at 0.05 level of significance

Conclusion:-

There is there is a difference between these proportions at α = 0.05

3 0

2 years ago

Luca had a large container of yogurt with 1 and StartFraction 5 Over 8 EndFraction pounds of yogurt left in it. If a serving of

lisov135 [29]

Answer:

4 servings

Step-by-step explanation:

Yogurt in the container = 1 5/8 pounds

A serving = 3/8 pounds

how many servings of yogurt are in the container?

Servings in the yogurt container = total yogurt in the container / yogurt per servings

= 1 5/8 ÷ 3/8

= 13/8 × 8/3

= 13/3

= 4.33

To the nearest whole serving

= 4 servings

7 0

2 years ago

What are the classifications of each system? Drag the answer into the box to match each system. {4x−2y=102x−y=5 {y=2x−3−2x+y=−5

siniylev [52]

The first one is. coincident
second. inconsistent
third. consistent independent

I just did this quiz (:

7 0

2 years ago

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