Find $P(z \lt 1.23)$, the area under the standard normal curve to the left of $z=1.23$

$0.8906$Find $P(-0.67 \lt z \lt 0)$, the area under the standard normal curve between $z=-0.67$ and $z=0$.

$0.2486$Find each of the following probabilities

- $P(z \gt 2.30)$
- $P(-0.33 \lt z \lt 1.45)$
- $P(z \gt -2.1)$
- $P(z \lt -3.5)$

(a) $0.0107$; (b) $0.5558$; (c) $0.9821$; (d) $0.0002$Find the number $b$ that satisfies:

- $P(z \lt b) = 0.9664$
- $P(z \lt b) = 0.3050$
- $P(z \gt b) = 0.0078$

(a) $1.830$; (b) $-0.5101$; (c) $2.4181$;Find $Q_1$, $Q_2$, and $Q_3$ on the standard normal distribution.

$Q_1 = -0.6745$, $Q_2 = 0$, $Q_3 = 0.6745$Find the 90th percentile on the standard normal distribution.

$1.2816$Find $P(-1 \lt z \lt 1)$ and $P(-2 \lt z \lt 2)$. Does your answer agree with the Empirical Rule?

$0.6827$ and $0.9545$, respectively. Yes, these are very close to what is suggested by the Empirical Rule.The average salary for first-year teachers is $\$27,989$. If the distribution is approximately normal with standard deviation $\$3250$,

what is the probability that a randomly selected first-year teacher makes between $\$20,000$ and $\$30,000$ each year?

what is the probability that a randomly selected first-year teacher has a salary less than $\$20,000$?

$\mu = 27989$ and $\sigma = 3250$. Finding $z_{20000} = (20000 - 27989)/3250 = -2.4582$, and $z_{30000} = (30000 - 27989)/3250 = 0.6188$, we calculate $P(20000 \lt x \lt 30000)$ as $P(-2.4581 \lt z \lt 0.6188) = 0.7250$.

Using the calculations from part (a), we have $P(x \lt 20000) = P(z \lt -2.4582) = 0.0070$.

The national average SAT score is 1019. If we assume a normal distribution with standard deviation 90,

what is the probability that a randomly selected score exceeds 1200?

what is the 90th percentile score?

$\mu = 1019$ and $\sigma = 90$. Finding $z_{1200} = (1200 - 1019)/90 = 2.011$, we calculate $P(x \gt 1200)$ as $P(z \gt 2.011) = 1 - P(z \lt 2.011) = 0.0222$

The area left of $z = 1.282$ is $0.90$, so the $90$th percentile score is $1019 + (1.282)(90) = 1134$.

The average time a person spends at the Barefoot Landing Seaquarium is 96 minutes. The standard deviation is 17 minutes. Assume the variable is normally distributed.

If a visitor is selected at random, find the probability that he or she will spend at least 120 minutes at the seaquarium.

If a visitor is selected at random, find the probability that he or she will spend at most 80 minutes at the sequarium.

Suggest a time for a bus to return to pick up a group of tourists

$0.0790$

$0.1733$

$3$ standard deviations away from the mean will catch almost all of the tourists, by the empirical rule, so pick them up $96+(3)(17) = 147$ minutes after dropping them off at the Barefoot Landing Seaquarium.

The average charitable contribution itemized per income tax return in Pennsylvania is $\$792$. Suppose that the distribution of contributions is normal with a standard deviation of $\$103$. Find the limits for the middle 50% of contributions.

The $z$-score for $Q_1$ is $-0.6745$, so $792 \pm (0.6745)(103)$ which yields the interval $(722.53, 861.47)$Is the following data set normal? $$\begin{array}{cccccccc} 3 & 58 & 5 & 65 & 17 & 48 & 52 & 75\\ 21 & 76 & 58 & 36 & 100 & 111 & 34 & 41\\ 23 & 44 & 33 & 50 & 13 & 18 & 7 & 12\\ 20 & 24 & 66 & 28 & 28 & 31 & & \end{array}$$

The histogram is unimodal, but appears skewed (although notably, it is not significantly so as $I = 0.7064 \lt 1$).