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3 years ago

Which is equivalent to P(z ≥ 1.4)?

Mathematics

2 answers:

sweet-ann [11.9K]3 years ago

4 0

Answer:

1-P(z<1.4)

Step-by-step explanation:

Trust me the answer is 1-P(z<1.4).

BTW the thing between 1 and P is a minus sign

frosja888 [35]3 years ago

3 0

Answer:

B. 1-P(z<1.4)

B. 26%

Step-by-step explanation:

You might be interested in

In a school 50% of learner's are younger than 10 , 1/20 are 10years old and 1/10 are older than 10 but younger than 12, the rema

frosja888 [35]

Answer:

1/20=10 years old

In a school 50% of the students are younger than 10, 1/20 are 10 years old and 1/10 are older than 10 but younger than 12, the remaining 70 students are 12 years or older. How many students are 10 years old?

50%= younger than 10

1/20=10 years old

1/10=older than 10 younger than 12

70=12 years or older

50%- 10/20

1/20-1/20

1/10-2/20

10/20+1/20+2/20=13/20

70 students =7/20

1/20 = 10 years old.

7/20 divided by 7 = 1/20

70 divided by 7=10

10 students are 10 years old

4 0

2 years ago

Read 2 more answers

Assume that the one-way commute time of an UoU student from his house to school is a normally distributed random variable which

Marizza181 [45]

Answer:

$n=(\frac{1.960(10)}{5})^2 =15.36 \approx 16$

So the answer for this case would be n=16 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=10$ represent the sample standard deviation

n represent the sample size

ME=5 represent the margin of error

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (2)

And on this case we have that ME =5 and we are interested in order to find the value of n, if we solve n from equation (2) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (3)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.96$ , replacing into formula (3) we got: