I do not understand the first calculation. We want the probability that the age is between $\frac{0.2}{3.7}$ standard deviation units below the mean and $\frac{1.3}{3.7}$ standard deviation units above the mean.
Using the table of the standard normal, we find that the probability that the age is $\frac{1.3}{3.7}\approx 0.351$ standard units above the mean or less is approximately $0.637$.
The probability that we are below the mean by $\frac{0.2}{3.7}$ or more standard deviation units is about $0.48$. So our required probability is about $0.637-0.48$, some distance from your number.
For the next problem, if we take $15$ ages and average them, the resulting random variable has normal distribution with mean $36.2$ and standard deviation $\frac{3.7}{\sqrt{15}}\approx 0.955$. Now one needs to essentially repeat the first calculation, but with $0.955$ replacing $3.7$. The probabilities will change dramatically.
Added: Let $X$ be normally distributed with mean $\mu$ and standard deviation $\sigma$. Then
$$\Pr(X\le a)=\Pr\left(Z\le \frac{a-\mu}{\sigma}\right),$$
where $Z$ is standard normal.
For reasonable positive values of $z$, $\Pr(Z\le z)$ can be looked up approximately in a table of the standard normal.
For negative $z$, we have to work a bit. If $z$ is negative, then $\Pr(Z\le z)=1-\Pr(Z\le |z|)$.
Tables of the standard normal are available on the web. Typically they are a single page, you can download one and print it. Such a table is still often found at the back of introductory statistics books.
But nowadays the reasonable thing is to let software do the work for you. There are many pieces of software, including spreadsheet programs, that have normal distribution calculations as a built in feature.
If you have independent events X, Y, the variance of these events is simply the sum of variances.
You can calculate the variance first and after that calculate the standard deviation. Eg.
$$Var(T) = Var(X_1) + Var(X_2) + Var(X_3) + Var(X_4)$$
$$Var(T) = 15^2 + 15^2 + 15^2 + 15 ^ 2$$
And standard deviation is the square root of this.
$$\sigma(T) = {\sqrt{Var(T)}} = \sqrt{900} = 30$$
The reason is that you cannot add standard deviations but you can add variances.
Best Answer
If you wanted to prove to yourself that fewer than 7 people recognize the brand name, you could employ a 'one sample t-test'.
First, lets set up two hypotheses; $H_o$ and $H_a$.
We interpret the alternative hypothesis $H_a$ as what the 'researcher' believes, which in this case is you.
So, as the researcher you believe that the means are different. Not one greater than the other, just different, or 'unusual'.
Therefore,
$$H_a: \mu_A\ne \mu_B $$
Next, we can formulate $H_o$. Here $H_o$, which is the null hypothesis, is just the opposite of the alternative hypothesis $H_a$. So, the opposite of 'not equal' is really just, 'equal'.
Therefore,
$$H_o: \mu_A=\mu_B $$
Before we get into the nitty-gritty, lets think about what showing that a mean of $7$ is significantly different from $10.3$ would mean.
If we showed that $7$ was significantly different from $10.3$, that would mean what? - It would mean that everything less than $7$ is also significantly different.
Now we can use a couple of formulas to determine if the mean of the sample, which we will take as $7$, is significantly different from the accepted mean of $10.3$.
The two formulas of interest will be;
$$t = t_{\alpha , n-1}$$ and,
$$t.s.v. = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}$$
Here '$t$' represents our 'critical value', while '$t.s.v.$' represents our 'test statistic value'. (Hopefully you're familiar with what each of the variables above mean, and you're able to substitute the values appropriately.)
Once you evaluate $t$ and $t.s.v.$, you want to check which one is greater - which leads us to the final step in answering the question - the decision rule.
Here, our decision rule will be, "If $t.s.v.>t$, accept $H_o$." Or in other words, the accepted sample mean ($10.3$) is not different to the new sample mean ($7$).