A distribution of measurements is relatively mound-shaped(normal distribution)with mean 50 and standard deviation 10. (Hint: use the Empirical Rule & show the figures) a. What proportion of the measurements will fall between 40 and 60?b. What proportion of the measurements will fall between 30 and 70? c. What proportion of the measurements will fall between 30 and 60?d. If a measurement is chosen at random from this distribution, what is the probability that it will begreater than 60?

a)

b)

c)

d)

Step-by-step explanation:

The empirical rule, also known as three-sigma rule or 68-95-99.7 rule, "is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)".

Let X the random variable who represent the measurements.

From the problem we have the mean and the standard deviation for the random variable X.

So we can assume

On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:

•The probability of obtain values within one deviation from the mean is 0.68

•The probability of obtain values within two deviation's from the mean is 0.95

•The probability of obtain values within three deviation's from the mean is 0.997

So we need values such that

Part a

We want this probability:

Using the empirical rule probabilities we can do this:

Part b

We want this probability:

Using the empirical rule probabilities we can do this:

Part c

We want this probability:

Using the empirical rule probabilities we can do this:

Part d

We want this probability:

Using the empirical rule probabilities and the complement rule we can do this:

is the anwser b. (-13)?

asnwer is (g o h o f)(x) = 2x - 6/x + 7

step-by-step explanation:

given,

f(x) = x - 3/x

g(x) = x + 3

h(x) = 2x + 1

we need to find : (g o h o f)(x)

we can write (g o h o f)(x) = g[ h{ f(x) } ]

so first we find h{f(x)}

h{f(x)} = f(x) + 3 = (x - 3/x) + 3

h{f(x)} = x - 3/x + 3

now we find g[ h{ f(x) } ].

g[ h{ f(x) } ] = g(x - 3/x + 3)       {plug h{f(x)} = x - 3/x + 3 in g[h{f(x)}] }

g(x - 3/x + 3) = 2 (x - 3/x + 3) + 1

                    = 2x - 6/x + 6 + 1

                    = 2x - 6/x + 7

hence we get (g o h o f)(x) = 2x - 6/x + 7