CAN SOMEONE PLEASE HELP ME WITH THESE!!!
2 answers:
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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 , 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:
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:
Compute the value of <em>z</em> as follows:
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:
Thus, the middle 30% of American men consume between 9.5 grams to 10.1 grams of sodium.