Question

11:44 AM Tue May 12 68% Q + : 0 Aren under a normal curve to the left of 2 where : * | 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 -0.8 0.2119 0.2090 0.2061 0.2033 0.2005 0.1977 0.1949 0.1922 0.1894 0.1867 -0.7 0.2420 0.2389 0.2358 0.2327 0.2296 0.2266 0.2236 0.2206 0.2177 0.2148 -0.6 0.2743 0.2709 0.2676 0.2643 0.2611 0.2578 0.2546 0.2514 0.2483 0.2451 -0.5 0.3085 0.3050 0.3015 0.2981 0.2946 0.2912 0.2877 0.2843 0.2810 0.2776 -0.4 0.3446 0.3409 0.3372 0.3336 0.3300 0.3264 0.3228 0.3192 0.3156 0.3121 -0.3 0.3821 0.3783 0.3745 0.3707 0.3669 0.3632 0.3594 0.3557 0.3520 0.3483 -0.2 0.4207 0.4168 0.4129 0.4090 0.4052 0.4013 0.3974 0.3936 0.3897 -0.1 0.4602 0.4562 0.4522 0.4483 0.4443 0.4404 0.4364 0.4325 0.4286 0.4247 0.0 0.5000 0.4960 0.4920 0.4880 0.4840 0.4801 0.4761 0.4721 0.4681 0.4641 0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359 0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753 0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141 0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517 0.4 0.6554 0.6591 0.6628 0.66640.6700 0.6736 0.6772 0.6808 0.6814 0.6879 0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224 0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549 0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.78230.7852 0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133 0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389 1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621 1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830 1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177 F(-2) = 1 - F(x)

Answer 1

a)

This is a normal distribution question with

$\\Mean (\mu)= 3.2 \\Standard\;Deviation (\sigma)= 0.5 \\Since\; we\; know\; that \\z_{ score } = \frac{x-\mu}{\sigma}$

P(2.9 < x < 3.6)=?

$\\ The\; z-score\; at\; x = 2.9 is, \\ z = \frac{2.9-3.2}{0.5} \\ z_1 = -0.6 \\ The\; z-score\; at\; x = 2.9 is, \\ z = \frac{3.6-3.2}{0.5} \\ z_2 = 0.8$

This implies that

P(2.9 < x < 3.6) = P(-0.6 < z < 0.8) = P(Z < 0.8) - P(Z < -0.6)

P(2.9 < x < 3.6) = 0.7881446014166034 - 0.2742531177500736

$P(2.9 < x < 3.6) = \textbf{0.5139}$

b)

$\mu=3$

Given in the question

P(X < 2.6) = 0.3085

This implies that

P(Z < -0.50) = 0.3085

$\\Since\; we\; know\; that \\z_{ score } = \frac{x-\mu}{\sigma} \\\\\\-0.50 =\frac{2.6-3}{\sigma} \\\\\sigma = 0.8$

PS: you have to refer z score table to find the final probabilities.
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