91Ó°ÊÓ

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 $P(\underset{}{\overset{}{∑}}x>170)$.

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

We have

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

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 $P\left(\underset{}{\overset{}{∑}}x<k_2\right)-0.85$

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

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$173.8393$.