Question

The grades on a statistics test are normally distributed with a mean of 62 and Q1=52. If the instructor wishes to assign B's or higher to the top 30% of the students in the class, what grade is required to get a B or higher? Please round your answer to two decimal places.

Answer 1

Answer:

To get a B or higher, you need to get a grade of at least 69.77.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that

Mean of 62, so $\mu = 62$

Q1 of 52 means that the z score of X = 52 has a pvalue of 0.25. Z has a pvalue of 0.25 between -0.67 and -0.68, so we use $Z = -0.675$

$Z = \frac{X - \mu}{\sigma}$

$-0.675 = \frac{52 - 62}{\sigma}$

$-0.675\sigma = -10$

Multiplying the equality by (-1), we have that:

$0.675\sigma = 10$

$\sigma = \frac{10}{0.675}$

$\sigma = 14.8$

If the instructor wishes to assign B's or higher to the top 30% of the students in the class, what grade is required to get a B or higher?

Those are the Z scores that have a pvalue higher than 0.70. So we have to find X when Z has a pvalue of 0.70. This is between 0.52 and 0.53, so we use $Z = 0.525$

$Z = \frac{X - \mu}{\sigma}$

$0.525 = \frac{X - 62}{14.8}$

$X - 62 = 14.8*0.525$

$X = 69.77$

To get a B or higher, you need to get a grade of at least 69.77.

Answer 2

Probabilities are used to determine the chances of an event.

To get a B or higher, a student needs to score at least 69.76

The given parameters are:

$\mathbf{\mu = 62}$ --- the mean

$\mathbf{x =Q_1 = 52}$ --- the first quartile

From the z-table,

The z score of the first quartile is:

$\mathbf{z = -0.67 }$

The z score formula is:

$\mathbf{z = \frac{x - \mu}{\sigma} }$

So, we have:

$\mathbf{-0.67 = \frac{52 - 62}{\sigma}}$

$\mathbf{-0.67 = \frac{- 10}{\sigma}}$

Rewrite as:

$\mathbf{\sigma= \frac{- 10}{-0.67 }}$

$\mathbf{\sigma= 14.93}$

To get 30% or above, it means that the p value is:

$\mathbf{p= 1 - 30\%}$

$\mathbf{p= 70\%}$

The z value at $\mathbf{ p= 70\% }$ is

$\mathbf{z = 0.52}$

Recall that

$\mathbf{z = \frac{x - \mu}{\sigma} }$

Substitute $\mathbf{\mu = 62}$, $\mathbf{z = 0.52}$ and $\mathbf{\sigma= 14.93}$

So, we have:

$\mathbf{0.52 = \frac{x - 62}{14.93}}$

Cross multiply

$\mathbf{0.52 \times 14.93= x - 62}$

$\mathbf{7.76 = x - 62}$

Make x the subject

$\mathbf{x = 7.76 + 62}$

$\mathbf{x = 69.76}$

Hence, to get a B or higher, a student needs to score at least 69.76

Read more about probabilities at:

https://brainly.com/question/16464000