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2 years ago

12

Please help and thank you

1 answer:

Vlada [557]2 years ago

3 0

Answer:

D

Step-by-step explanation:

Lisa wants to spend less than 200 dollars, and has already spent 125.74. She also wants to buy as many shirts that cost 20.35 dollars as possible.

Each shirt costs 20.35, and Lisa want to buy as many shirts as possible. Having "shirts" = x, it'd be 20.35x + 125.74 < 200.

20.35 = How much each shirt costs

X = How many shirts she can buy

125.74 = How much she has already spent

200 = Her limit on how much she can spend.

Hope this helps!

Please correct me if I've done anything wrong, and feel free to elaborate if you feel like I haven't explained something clearly!

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A normal distribution has a standard deviation equal to 39. What is the mean of this normal distribution if the probability of s

Naily [24]

Answer:

The mean is $\mu = 131$

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\sigma = 39$

What is the mean of this normal distribution if the probability of scoring above x = 209 is 0.0228?

This means that when X = 209, Z has a pvalue of 1-0.0228 = 0.9772. So when X = 209, Z = 2.

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

$2 = \frac{209 - \mu}{39}$

$209 - \mu = 2*39$

$\mu = 209 - 78$

$\mu = 131$

The mean is $\mu = 131$

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2 years ago

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(Ross 5.15) If X is a normal random variable with parameters µ " 10 and σ 2 " 36, compute (a) PpX ą 5q (b) Pp4 ă X ă 16q (c) PpX

pochemuha

Answer:

(a) 0.7967

(b) 0.6826

(c) 0.3707

(d) 0.9525

(e) 0.1587

Step-by-step explanation:

The random variable <em>X</em> follows a Normal distribution with mean <em>μ</em> = 10 and variance <em>σ</em>² = 36.

(a)

Compute the value of P (X > 5) as follows:

$P(X>5)=P(\frac{x-\mu}{\sigma}>\frac{5-10}{\sqrt{36}})\\=P(Z>-0.833)\\=P(Z$

Thus, the value of P (X > 5) is 0.7967.

(b)

Compute the value of P (4 < X < 16) as follows:

$P(4$

Thus, the value of P (4 < X < 16) is 0.6826.

(c)

Compute the value of P (X < 8) as follows:

$P(X$

Thus, the value of P (X < 8) is 0.3707.

(d)

Compute the value of P (X < 20) as follows:

$P(X$

Thus, the value of P (X < 20) is 0.9525.

(e)

Compute the value of P (X > 16) as follows:

$P(X>16)=P(\frac{x-\mu}{\sigma}>\frac{16-10}{\sqrt{36}})\\=P(Z>1)\\=1-P(Z$

Thus, the value of P (X > 16) is 0.1587.

**Use a <em>z</em>-table for the probabilities.

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