Question

The Massachusetts State Lottery averages, on a weekly basis, a profit of 10.0 million dollars. The variability, as measured by the variance statistic is 6.25 million dollars squared.
If it is known that the weekly profits is distributed normally, what are the chances that, in a given week, the profits will be between 8 and 10.5 million dollars?

Part of Φ(z) table
z +0.00 +0.01 +0.02 +0.03 +0.04 +0.05 +0.06 +0.07 +0.08 +0.09
0.0 0.50000 0.50399 0.50798 0.51197 0.51595 0.51994 0.52392 0.52790 0.53188 0.53586
0.1 0.53983 0.54380 0.54776 0.55172 0.55567 0.55966 0.56360 0.56749 0.57142 0.57535
0.2 0.57926 0.58317 0.58706 0.59095 0.59483 0.59871 0.60257 0.60642 0.61026 0.61409
0.3 0.61791 0.62172 0.62552 0.62930 0.63307 0.63683 0.64058 0.64431 0.64803 0.65173
0.4 0.65542 0.65910 0.66276 0.66640 0.67003 0.67364 0.67724 0.68082 0.68439 0.68793
0.5 0.69146 0.69497 0.69847 0.70194 0.70540 0.70884 0.71226 0.71566 0.71904 0.72240
0.6 0.72575 0.72907 0.73237 0.73565 0.73891 0.74215 0.74537 0.74857 0.75175 0.75490
0.7 0.75804 0.76115 0.76424 0.76730 0.77035 0.77337 0.77637 0.77935 0.78230 0.78524
0.8 0.78814 0.79103 0.79389 0.79673 0.79955 0.80234 0.80511 0.80785 0.81057 0.81327
0.9 0.81594 0.81859 0.82121 0.82381 0.82639 0.82894 0.83147 0.83398 0.83646 0.83891

Answer 1

Answer:

Probability that in a given week, the profits will be between 8 and 10.5 million dollars is 0.3674.

Step-by-step explanation:

We are given that the Massachusetts State Lottery averages, on a weekly basis, a profit of 10.0 million dollars. The variability, as measured by the variance statistic is 6.25 million dollars squared.

Also, it is known that the weekly profits is distributed normally.

Firstly, Let X = weekly profits

The z score probability distribution for is given by;

           Z = $\frac{ X - \mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean profit = 10 million dollars

           $\sigma$ = standard deviation = $\sqrt{variance}$ = $\sqrt{6.25}$ = 2.5 million dollars

Probability that, in a given week, the profits will be between 8 and 10.5 million dollars is given by = P(8 < X < 10.5) = P(X < 10.5) - P(X $\leq$ 8)

  P(X < 10.5) = P( $\frac{ X - \mu}{\sigma}$ < $\frac{10.5-10}{2.5}$ ) = P(Z < 0.2) = 0.57926

  P(X $\leq$ 8) = P( $\frac{ X - \mu}{\sigma}$ $\leq$ $\frac{8-10}{2.5}$ ) = P(Z $\leq$ -0.8) = 1 - P(Z < 0.8)

                                            = 1 - 0.78814 = 0.21186

Therefore, P(8 < X < 10.5) = 0.57926 - 0.21186 = 0.3674

Hence, the chances that, in a given week, the profits will be between 8 and 10.5 million dollars is 0.3674.