I-Ready

Suppose that the walking step lengths of adult males are normally distributed with a mean of 2.5 feet and a standard deviation o

Mumz [18]

Answer:

0% probability that the mean of the sample taken is less than 2.2 feet.

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.

Mean of 2.5 feet and a standard deviation of 0.2 feet.

This means that $\mu = 2.5, \sigma = 0.2$

Sample of 41

This means that $n = 41, s = \frac{0.2}{\sqrt{41}}$

Find the probability that the mean of the sample taken is less than 2.2 feet.

This is the p-value of Z when X = 2.2 So

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

By the Central Limit Theorem

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

$Z = \frac{2.2 - 2.5}{\frac{0.2}{\sqrt{41}}}$

$Z = -9.6$

$Z = -9.6$ has a p-value of 0.

0% probability that the mean of the sample taken is less than 2.2 feet.

6 0

3 years ago

Ali and his father went on a trip to Jazan. Ali drove 52 miles less than his

kenny6666 [7]

Answer:

90

Step-by-step explanation:

divide 292 by 2

subtract 56 to get Ali's and add 56 to his Dad's

I could be wrong it's 6 am

7 0

2 years ago

Please help me with the below question.

VMariaS [17]

By letting

$y = \displaystyle \sum_{n=0}^\infty c_n x^{n+r}$

we get derivatives

$y' = \displaystyle \sum_{n=0}^\infty (n+r) c_n x^{n+r-1}$

$y'' = \displaystyle \sum_{n=0}^\infty (n+r) (n+r-1) c_n x^{n+r-2}$

a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to

$5r(r-1) c_0 x^{r-1} + \displaystyle \sum_{n=1}^\infty \bigg( (n+r+1) c_n + (n + r + 1) (5n + 5r + 1) c_{n+1} \bigg) x^{n+r} = 0$

Examine the lowest degree term $\left(x^{r-1}\right)$ , which gives rise to the indicial equation,

$5r (r - 1) + r = 0 \implies 5r^2 - 4r = r (5r - 4) = 0$

with roots at r = 0 and r = 4/5.

b) The recurrence for the coefficients $c_k$ is

$(k+r+1) c_k + (k + r + 1) (5k + 5r + 1) c_{k+1} = 0 \implies c_{k+1} = -\dfrac{c_k}{5k+5r+1}$

so that with r = 4/5, the coefficients are governed by

$c_{k+1} = -\dfrac{c_k}{5k+5} \implies \boxed{g(k) = -\dfrac1{5k+5}}$

c) Starting with $c_0=1$ , we find

$c_1 = -\dfrac{c_0}5 = -\dfrac15$

$c_2 = -\dfrac{c_1}{10} = \dfrac1{50}$

so that the first three terms of the solution are

$\displaystyle \sum_{n=0}^2 c_n x^{n + 4/5} = \boxed{x^{4/5} - \dfrac15 x^{9/5} + \frac1{50} x^{13/5}}$

4 0

2 years ago

2 1/5 × 5/6<br> (its a fraction) <br> PLSSS i need it now its a test

Soloha48 [4]

Answer:

$\frac{11}{6} \ \ or \ 1\frac{5}{6}$

Step-by-step explanation:

$2\frac{1}{5} \times \frac{5}{6} \\\\ = \frac{11}{5} \times \frac{5}{6} \\\\=\frac{11}{6}\\\\=1 \frac{5}{6}$

6 0

2 years ago

Read 2 more answers

Which inequality represents this sentence?

olga_2 [115]

6 + 2 < 9 + 3...u r correct

5 0

3 years ago

Read 2 more answers