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2 years ago

For a normal distribution,

Mathematics

1 answer:

Anni [7]2 years ago

7 0

Answer:

1.282, -0.5244,...

Step-by-step explanation:

We know that Z is a standard normal variate with mean =0, and std dev =1

a) z-score separates the highest 10% from the rest of the scores

= 90th percentile = 1.282

b) z-score separates the lowest 30% from the rest of the scores

= 30th percentile = -0.5244

c) z-score separates the lowest 40% from the rest of the scores

= 40th percentile = -0.2533

d) z-score separates the lowest 20% from the rest of the score

=20th percentile = -0.8416

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What is 19/5 in mixed number?

nlexa [21]

3 4/5 is the mixed number. I hope this helps you :]

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3 years ago

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Write the equation of the line in fully simplified slope intercept form.

umka2103 [35]

Answer:

y = -1/3 x - 7

Step-by-step explanation:

First, the intercept form is in y = ax + b, where b is the intercept at y-axis which is -7, then pick any point in the line, in the case I chose (3,-8), then substitute it into the equation, -8 = 3a - 7, then solve for a, get 3a = -1, so a = -1/3 which is the slope.

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2 years ago

ABC and AED are straight lines. BE and CD are parallel. AC = 12.3cm AB = 8.2cm BE = 3.8cm a) Work out length CD. AD = 9.15cm

elena-14-01-66 [18.8K]

Answer:

15.33cm

Step-by-step explanation:

when you try to draw out this figure, satisfying the fact that Ad and Ac are both straight lines. you find the ultimately it forms a right-angled triangle with sides AC and AD and hypotenuse CD.

Now, you may use Pythagoras theorem to solve:

AC^2 + AD^2 = CD^2

12.3^2 + 9.15^2 = CD^2

CD^2 = 235.0125

CD = square root of 235.0125 = 15.33cm

Hope this helps

6 0

3 years ago

A certain flight arrives on time 8484 percent of the time. Suppose 143143 flights are randomly selected. Use the normal approxim

Troyanec [42]

Answer:

a) P(x=108)=0.0020

b) P(x≥108)=0.9980

c) P(x<124)=0.7794

d) P(124≤x≤128)=0.1925

Step-by-step explanation:

We know the population proportion, that is p=0.84.

We take a sample of size n=143.

We will use the normal approximation to the binomial distribution to model this problem.

The mean and standard deviation of the normal approximation to the binomial distribution will be:

$\mu=np=143*0.84=120.12\\\\\sigma=\sqrt{np(1-p)}=\sqrt{143*0.84*0.16}=\sqrt{19.22}=4.38$

a) We have to calculate the probability that exactly 108 flights are on time.

As the normal distribution considers the random variable to be continous, we have to apply the continuity correction factor.

In this case, the probability of 108 flights on time can be calculated as P(107.5<x<108.5):

$P(x=108)=P(107.5$

b) Now we have to calculate that at least 108 flights are on time.

As the probability includes 108, the continuity factor will indicates that we calculate P(x>107.5). The z-value for x=107.5 has been already calculated in point a:

$P(x\geq108)=P(x>107.5)=P(z>-2.88)=0.9980$

c) We have to calculate the probability that fewer than 124 flights are on time. According to the continuity factor, we have to calculate the probability P(x<123.5), as the flight number 124 is not included in the interval.

$P(x$