A normal distribution has a mean of 67 with a standard deviation of 8. Find the probability that a randomly selected x-value is at least 75

P(X≥75)= 0.22663

Step-by-step explanation:

Hello!

The variable has a normal distribution with mean μ= 67 and a standard deviation δ= 8.

X~N(μ;δ²)

You need to find the probability that a randomly selected X value is at least 75, i.e. 75 or more, symbolically:

P(X≥75)

To calculate this probability is best to work using the standard normal distribution. This distribution is derived from the normal distribution. Considering a random variable X with normal distribution, mean μ and variance δ², the variable Z =(X-μ)/δ ~N(0;1) is determined.

The standard normal distribution is tabulated. Any value of any random variable X with normal distribution can be "converted" by subtracting the variable from its mean and dividing it by its standard deviation.

Using the formula and the values of the parameters of the population you have to standardize the value of X

Z= (X-μ)/δ = (75 - 67)/ 8= 0.75

Now you can express the probability under the standard normal distribution:

P(X≥75) = P(Z≥0.75)

The Z-tables show probability values P(Z≤z₁₋α)= 1 - α, then:

P(Z≥0.75)= 1 - P(Z<0.75)= 1 - 0.77337= 0.22663.

I hope it helps!

the answer is; biodiversity resulting from few ancestors

adaptive radiation occurs when a population is subjected to a large natural environment with spatial ecological variations. the parts of the population in the different ecologies begin to adapt to their local environments. due to non-random mating across the large populations, this causes the different populations from the large population to speciate within their local environments. they, therefore, diverge to different species sharing a common ancestor.