We are regarding a normal distribution with a mean of 35 and a standard deviation of 6, i.e., $\mu$ = 35 and $\sigma$ = 6. We know that the probability density function for a normal distribution with a mean of $\mu$ and a standard deviation of $\sigma$ is given by

$f(x) = \frac{1}{\sqrt{2\pi}\sigma}\exp[-\frac{(x-\mu)^{2}}{2\sigma^{2}}]$

in this case we have

$f(x) = \frac{1}{\sqrt{2\pi}6}\exp[-\frac{(x-35)^{2}}{2(6^{2})}]$

Let $X$ be the random variable that represents a row score, we find the values we are seeking in the following way

a) we are looking for a number $x_{0}$ such that

$P(X\leq x_{0})$ = $\int\limits^{x_{0}}_{-\infty} {f(x)} \, dx = 0.65$ , this number is $x_{0}$ =37.31

you can find this answer using the R statistical programming languange and the instruction qnorm(0.65, mean = 35, sd = 6)

b) we are looking for a number $x_{1}$ such that

$P(X\leq x_{1})$ = $\int\limits^{x_{1}}_{-\infty} {f(x)} \, dx = 0.9$ , this number is $x_{1}$ =42.70

you can find this answer using the R statistical programming languange and the instruction qnorm(0.9, mean = 35, sd = 6)

c) we find this probability as

$P(28\leq X\leq 38)$ = $\int\limits^{38}_{28} {f(x)} \, dx = 0.57$

you can find this answer using the R statistical programming languange and the instruction pnorm(38, mean = 35, sd = 6) -pnorm(28, mean = 35, sd = 6)

d) we find this probability as

$P(41\leq X\leq 44)$ = $\int\limits^{44}_{41} {f(x)} \, dx = 0.09$

you can find this answer using the R statistical programming languange and the instruction pnorm(44, mean = 35, sd = 6) -pnorm(41, mean = 35, sd = 6)

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