Which referent could you use for 1 m?

The time for a visitor to read health instructions on a Web site is approximately normally distributed with a mean of 10 minutes

klio [65]

Answer:

a) The mean is 10 and the variance is 0.0625.

b) 0.6826 = 68.26% probability that the mean time of the visitors is within 15 seconds of 10 minutes.

c) 10.58 minutes.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

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.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Normally distributed with a mean of 10 minutes and a standard deviation of 2 minutes.

This means that $\mu = 10, \sigma = 2$

Suppose 64 visitors independently view the site.

This means that $n = 64, = \frac{2}{\sqrt{64}} = 0.25$

a. The expected value and the variance of the mean time of the visitors.

Using the Central Limit Theorem, mean of 10 and variance of (0.25)^2 = 0.0625.

b. The probability that the mean time of the visitors is within 15 seconds of 10 minutes.

15 seconds = 15/60 = 0.25 minutes, so between 9.75 and 10.25 seconds, which is the p-value of Z when X = 10.25 subtracted by the p-value of Z when X = 9.75.

X = 10.25

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

By the Central Limit Theorem

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

$Z = \frac{10.25 - 10}{0.25}$

$Z = 1$

$Z = 1$ has a p-value of 0.8413.

X = 9.75

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

$Z = \frac{9.75 - 10}{0.25}$

$Z = -1$

$Z = -1$ has a p-value of 0.1587.

0.8413 - 0.1587 = 0.6826.

0.6826 = 68.26% probability that the mean time of the visitors is within 15 seconds of 10 minutes.

c. The value exceeded by the mean time of the visitors with probability 0.01.

The 100 - 1 = 99th percentile, which is X when Z has a p-value of 0.99, so X when Z = 2.327.

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

$2.327 = \frac{X - 10}{0.25}$

$X - 10 = 2.327*0.25$

$X = 10.58$

So 10.58 minutes.

6 0

2 years ago

Find the quotient of 1 4 ÷ 3 8 in simplest form

Otrada [13]

Is that suppose to be 14/38?

4 0

2 years ago

Read 2 more answers

Answer true and false please thanks

Aleksandr [31]

Option a is true and others are false

6 0

2 years ago

Divide x^3+2 by x+1 long division ppleeease! :)

anzhelika [568]

Long division: (x³ + 2) ÷ (x + 1)

 x² – x + 1 
x³ + 0x² + 0x + 2 | x + 1
– x³ – x² ⋮ ⋮
– x² + 0x ⋮
+ x² + x ⋮
+ x + 2
 – x – 1
+ 1

 Quotient: Q(x) = x² – x – 1;

Remainder: R(x) = + 1.

I hope this helps. =)

3 0

3 years ago

Elisa needs 40 shirts for a charity. One store is selling shirts for $6.95 each. Another store says that the total for al 40 shi

xxTIMURxx [149]

Well, let us figure out what the first store will cost.
40*6.95=278
Now for the second store
325*0.8=260

So the first store is $278, while the second store is $260, so Elisa should buy shirts from the second store

7 0

3 years ago