In a recent​ year, the total scores for a certain standardized test were normally​ distributed, with a mean of 500 and a standard deviation of 10.4. A) Find the probability that a randomly selected medical student who took the test had a total score that was less than 484. The probability that a randomly selected medical student who took the test had a total score that was less than 484 is:.B) Find the probability that a randomly selected study participant's response was between 4 and 6 The probability that a randomly selected study participant's response was between 4 and 6 is:.C) Find the probability that a randomly selected study participant's response was more than 8. The probability that a randomly selected study participant's response was more than 8 is:.

A) The probability that a randomly selected medical student who took the test had a total score that was less than 484 = 0.06178

B) The probability that a randomly selected study participant's response was between 504 and 516 = 0.29019

C) The probability that a randomly selected study participant's response was more than 528 = 0.00357

D) Option D is correct.

Only the event in (c) is unusual as its probability is less than 0.05.

Step-by-step explanation:

The b and c parts of the question are not complete.

B) Find the probability that a randomly selected study participant's response was between 504 and 516

C) Find the probability that a randomly selected study participant's response was more than 528.

D) Identify any unusual event amongst the three events in A, B and C. Explain the reasoning.

a) None.

b) Events A and B.

C) Event A

D) Event C

Solution

This is a normal distribution problem with

Mean = μ = 500

Standard deviation = σ = 10.4

A) Probability that a randomly selected medical student who took the test had a total score that was less than 484 = P(x < 484)

We first normalize or standardize 484

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (484 - 500)/10.4 = - 1.54

To determine the required probability

P(x < 484) = P(z < -1.54)

We'll use data from the normal distribution table for these probabilities

P(x < 484) = P(z < -1.54) = 0.06178

B) Probability that a randomly selected study participant's response was between 504 and 516 = P(504 ≤ x ≤ 516)

We normalize or standardize 504 and 516

For 504

z = (x - μ)/σ = (504 - 500)/10.4 = 0.38

For 516

z = (x - μ)/σ = (516 - 500)/10.4 = 1.54

To determine the required probability

P(504 ≤ x ≤ 516) = P(0.38 ≤ z ≤ 1.54)

We'll use data from the normal distribution table for these probabilities

P(504 ≤ x ≤ 516) = P(0.38 ≤ z ≤ 1.54)

= P(z ≤ 1.54) - P(z ≤ 0.38)

= 0.93822 - 0.64803

= 0.29019

C) Probability that a randomly selected study participant's response was more than 528 = P(x > 528)

We first normalize or standardize 528

z = (x - μ)/σ = (528 - 500)/10.4 = 2.69

To determine the required probability

P(x > 528) = P(z > 2.69)

We'll use data from the normal distribution table for these probabilities

PP(x > 528) = P(z > 2.69) = 1 - P(z ≤ 2.69)

= 1 - 0.99643

= 0.00357

D) Only the event in (c) is unusual as its probability is less than 0.05.

Hope this Helps!!!

225x²+4=0

225x²=0-4

225x²=-4

x²=-4/225

x=±√-4/225