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Rashid [163]
2 years ago
8

HELP

Mathematics
2 answers:
Lesechka [4]2 years ago
6 0

Answer: the answer to dat is 12

Step-by-step explanation: because .20 x 60 = 12

uranmaximum [27]2 years ago
4 0
20% of 60 is 12

60/300 = 20%

The friend is incorrect because that is 20% of 300 not 60
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The average American man consumes 9.8 grams of sodium each day. Suppose that the sodium consumption of American men is normally
Alex Ar [27]

Answer:

(a) The distribution of <em>X</em> is <em>N</em> (9.8, 0.8²).

(b) The probability that an American consumes between 8.8 and 9.9 grams of sodium per day is 0.4461.

(c) The middle 30% of American men consume between 9.5 grams to 10.1 grams of sodium.

Step-by-step explanation:

The random variable <em>X</em> is defined as the amount of sodium consumed.

The random variable <em>X</em> has an average value of, <em>μ</em> = 9.8 grams.

The standard deviation of <em>X</em> is, <em>σ</em> = 0.8 grams.

(a)

It is provided that the sodium consumption of American men is normally distributed.

The random variable <em>X</em> follows a normal distribution with parameters <em>μ</em> = 9.8 grams and <em>σ</em> = 0.8 grams.

Thus, the distribution of <em>X</em> is <em>N</em> (9.8, 0.8²).

(b)

If X ~ N (µ, σ²), then Z=\frac{X-\mu}{\sigma}, is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z ~ N (0, 1).

To compute the probability of  Normal distribution it is better to first convert the raw score (<em>X</em>) to <em>z</em>-scores.

Compute the probability that an American consumes between 8.8 and 9.9 grams of sodium per day as follows:

P(8.8

                           =P(-1.25

Thus, the probability that an American consumes between 8.8 and 9.9 grams of sodium per day is 0.4461.

(c)

The probability representing the middle 30% of American men consuming sodium between two weights is:

P(x_{1}

Compute the value of <em>z</em> as follows:

P(-z

The value of <em>z</em> for P (Z < z) = 0.65 is 0.39.

Compute the value of <em>x</em>₁ and <em>x</em>₂ as follows:

-z=\frac{x_{1}-\mu}{\sigma}\\-0.39=\frac{x_{1}-9.8}{0.8}\\x_{1}=9.8-(0.39\times 0.8)\\x_{1}=9.488\\x_{1}\approx9.5     z=\frac{x_{2}-\mu}{\sigma}\\0.39=\frac{x_{1}-9.8}{0.8}\\x_{1}=9.8+(0.39\times 0.8)\\x_{1}=10.112\\x_{1}\approx10.1

Thus, the middle 30% of American men consume between 9.5 grams to 10.1 grams of sodium.

4 0
3 years ago
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