Use the information in Example 7.8, but use a sample size of 55 to answer the following questions. a. Find P( x ¯ < 7).

b. Find P(Σx > 170).

c. Find the 80th percentile for the mean of 55 scores.

d. Find the 85th percentile for the sum of 55 scores.

We have given that the sample size of 55.

We need to find$P(\overline{x}<7)-1$.

$P(\overline{x}<7)$

We have

normalcdf$\left(1,7,3,\frac{1.15}{\sqrt{55}}\right)-1$

The probability that the mean stress score is less than $7$is $1$.

Normal Distribution:

We have given that the sample size of $55$.

We need to find that $P(\sum _{}^{}x>170)$.

$P(\overline{x}>170)$

We have

normalcdf$(170,1E99,55\times 3\sqrt{55}\times 1.15)$

The probability that the total of $55$scores is more than $170$is $0.2788$.

Normal Distribution:

We have given that the sample size of $55$.

We need to find that the${80}^{th}$percentile.

Let $k_1-$the ${80}^{th}$percentile

We need to find $P\{\overline{x}<k_1)-0.80:$

invNorm$\left(0.80,3,\frac{1.15}{\sqrt{55}}\right)-3.1\text{'}3$

The ${80}^{th}$percentile for the mean of $55$score is about $3.13$.

This tells us about$80\%$of all the means of$55$stress scores are at most$3.13$and that$20\%$are at least$3.13$.

We heve given that the sample size of $55$.

We need to find that${85}^{th}$percentile.

Let $k_2-$the ${85}^{th}$percentile

We need to find $P\left(\sum _{}^{}x<k_2\right)-0.85$

invNorm$(0.85,55\times 3,\sqrt{55}\times 1.15)-173.8393$.

The ${85}^{th}$percentile for the sum of $55$scores is about $173.8393$.

This tells us that ${85}^{th}$of all the sums of$55$scores are no more than$173.8393$and$15\%$are no less than$173.8393$.