What does a multiplier/divider outside the function do to the graph?

1 answer:

7 0

If the coefficient squared is greater than one (multiplication), it stretches the graph vertically by the factor of the coefficient.

If the coefficient squared is less than one (division), it compresses the graph vertically by the factor of the coefficient.

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Jill’s bowling scores are approximately normally distributed with mean 170 and standard deviation 20, while Jack’s scores are ap

miss Akunina [59]

Answer:

a) The probability of Jack scoring higher is 0.3446

b) They probability of them scoring above 350 is 0.2119

Step-by-step explanation:

Lets call X the random variable that determines Jill's bowling score and Y the random variable that determines jack's. We have

$X \simeq N(170,400)\\Y \simeq N(160,225)$

Note that we are considering the variance on the second entry, the square of the standard deviation.

If we have two independent Normal distributed random variables, then their sum is also normally distributed. If fact, we have this formulas:

$N(\lambda_1, \sigma^2_1) + N(\lambda_2, \sigma^2_2) = N(\lambda_1 + \lambda_2,\sigma^2_1 + \sigma^2_2) \\r* N(\lambda_1, \sigma^2_1) = N(r\lambda_1,r^2\sigma^2_1)$

for independent distributions $N(\lambda_1, \sigma^2_1)$ , $N(\lambda_2, \sigma^2_2)$ , and a real number r.

a) We define Z to be Y-X. We want to know the probability of Z being greater than 0. We have

$Z = Y-X = N(160,225) - N(170,400) = N(160,225) + (N(-170,(-1)^2 * 400) = N(-10,625)$

So Z is a normal random variable with mean equal to -10 and vriance equal to 625. The standard deviation of Z is √625 = 25.

Lets work with the standarization of Z, which we will call W. $W = (Z-\mu)/\sigma$ = (Z+10)/25. W has Normal distribution with mean 0 and standard deviation 1. We have

P(Z > 0) = P( (Z+10)/25 > (0+10)/25) = P(W > 0.4)

To calculate that, we will use the <em>known </em>values of the cummulative distribution function Φ of the standard normal distribution. For a real number k, P(W < k) = Φ(k). You can find those values in the Pdf I appended below.

Since Φ is a cummulative distribution function, we have P(W > 0.4) = 1- Φ(0.4)

That value of Φ(0.4) can be obtained by looking at the table, it is 0.6554. Therefore P(W > 0.4) = 1-0.6554 = 0.3446

As a result, The probability of Jack's score being higher is 0.3446. As you may expect, since Jack is expected to score less that Jill, the probability of him scoring higher is lesser than 0.5.

b) Now we define Z to be X+Y Since X and Y are independent Normal variables with mean 160 and 170 respectively, then Z has mean 330. And the variance of Z is equal to the sum of the variances of X and Y, that is, 625. Hence Z is Normally distributed with mean 330 and standard deviation rqual to 25 (the square root of 625).

We want to know the probability of Z being greater that 350, for that we standarized Z. We call W the standarization. W is s standard normal distributed random variable, and it is obtained from Z by removing its mean 330 and dividing by its standard deviation 25.

P(Z > 350) = P((Z - 330)/25 > (350-330)/25) = P(W > 0.8) = 1-Φ(0.8)

The last equality comes from the fact that Φ is a cummulative distribution function. The value of Φ(0.8) by looking at the table is 0.7881, therefore P(X+Y > 350) = 1 - Φ(0.8) = 0.2119.

As you may expect, this probability is pretty low because the mean value of the sum of their combined scores is quite below 350.

I hope this works for you!

Download pdf

6 0

4 years ago

Im so confused pls help!

Digiron [165]

$-\frac{1}{2} *(-\frac{5}{6} )+(-\frac{1}{2} )*\frac{1}{3} \\=\frac{5}{12} -\frac{1}{6} \\=\frac{5}{12} -\frac{2}{12} \\=\frac{3}{12} \\\frac{1}{4}$