Question

A researcher was interested in seeing how many names a class of 38 students could remember after playing a name game After playing the name game, the students were asked to recall as many first names of fellow students as possible. The mean number of names recalled was 19.41 with a standard deviation of 3.17. Use this information to solve the following problem. What proportion of the students recalled more than 15 names?

Answer 1

Answer:

Proportion of the students recalled more than 15 names is 91.77%.

Step-by-step explanation:

We are given that a researcher was interested in seeing how many names a class of 38 students could remember after playing a name game After playing the name game, the students were asked to recall as many first names of fellow students as possible.

The mean number of names recalled was 19.41 with a standard deviation of 3.17.

Let X = number of names recalled

SO, X ~ N($\mu = 19.41,\sigma^{2} = 3.17^{2}$)

The z-score probability distribution is given by ;

                  Z = $\frac{X-\mu}{\sigma} } }$ ~ N(0,1)

where, $\mu$ = mean number of names recalled = 19.41

            $\sigma$ = standard deviation = 3.17

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, proportion of the students recalled more than 15 names is given by = P(X > 15 names)

     P(X > 15) = P( $\frac{X-\mu}{{\sigma} } }$ > $\frac{15-19.41}{3.17} }$ ) = P(Z > -1.39)

                                                      = P(Z < 1.39) = 0.9177  {using z table}

Therefore, proportion of the students recalled more than 15 names is 91.77%.